Shared Qs (015_linear)

1. 
   

   Question

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   Below, a specific arithmetic sequence, with a first element equal 20 and common difference equal 9, is defined recursively, where \(n\) is any whole number larger than 1. \[a_{[1]} = 20\] \[d = 9\] \[a_{[n]} = a_{[n-1]}+d\] Using that definition we can find the first few terms. \[a_{[2]} ~=~ a_{[1]}+d ~=~ 20+9 ~=~ 29 \] \[a_{[3]} ~=~ a_{[2]}+d ~=~ 29+9 ~=~ 38 \] \[a_{[4]} ~=~ a_{[3]}+d ~=~ 38+9 ~=~ 47 \] \[a_{[5]} ~=~ a_{[4]}+d ~=~ 47+9 ~=~ 56 \]

   We say the first element is 20, the second element is 29, the third element is 38, the fourth element is 47, and the fifth element is 56.

   The index is the address, so \(a_{[2]}=29\) says that the term with an index of 2 has a value of 29.

   To find the value of an element with a large index, it is useful to know the explicit formula:

   \[a_{[n]} ~~=~~ a_{[1]}~+~(n-1)\cdot d \]

   Find \(a_{[68]}\): the value of the element with an index equaling 68.

   
   

   Solution

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   Use the explicit formula.

   \[a_{[68]} = 20+(68-1)\cdot 9 \] \[a_{[68]} = 623 \]

   
   
2. 
   

   Question

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   Below, a specific arithmetic sequence, with a first element equal 30 and common difference equal 10, is defined recursively, where \(n\) is any whole number larger than 1. \[a_{[1]} = 30\] \[d = 10\] \[a_{[n]} = a_{[n-1]}+d\] Using that definition we can find the first few terms. \[a_{[2]} ~=~ a_{[1]}+d ~=~ 30+10 ~=~ 40 \] \[a_{[3]} ~=~ a_{[2]}+d ~=~ 40+10 ~=~ 50 \] \[a_{[4]} ~=~ a_{[3]}+d ~=~ 50+10 ~=~ 60 \] \[a_{[5]} ~=~ a_{[4]}+d ~=~ 60+10 ~=~ 70 \]

   We say the first element is 30, the second element is 40, the third element is 50, the fourth element is 60, and the fifth element is 70.

   The index is the address, so \(a_{[2]}=40\) says that the term with an index of 2 has a value of 40.

   To find the value of an element with a large index, it is useful to know the explicit formula:

   \[a_{[n]} ~~=~~ a_{[1]}~+~(n-1)\cdot d \]

   Find \(a_{[66]}\): the value of the element with an index equaling 66.

   
   

   Solution

   Show solution

   Use the explicit formula.

   \[a_{[66]} = 30+(66-1)\cdot 10 \] \[a_{[66]} = 680 \]

   
   
3. 
   

   Question

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   Below, a specific arithmetic sequence, with a first element equal 22 and common difference equal 3, is defined recursively, where \(n\) is any whole number larger than 1. \[a_{[1]} = 22\] \[d = 3\] \[a_{[n]} = a_{[n-1]}+d\] Using that definition we can find the first few terms. \[a_{[2]} ~=~ a_{[1]}+d ~=~ 22+3 ~=~ 25 \] \[a_{[3]} ~=~ a_{[2]}+d ~=~ 25+3 ~=~ 28 \] \[a_{[4]} ~=~ a_{[3]}+d ~=~ 28+3 ~=~ 31 \] \[a_{[5]} ~=~ a_{[4]}+d ~=~ 31+3 ~=~ 34 \]

   We say the first element is 22, the second element is 25, the third element is 28, the fourth element is 31, and the fifth element is 34.

   The index is the address, so \(a_{[2]}=25\) says that the term with an index of 2 has a value of 25.

   To find the value of an element with a large index, it is useful to know the explicit formula:

   \[a_{[n]} ~~=~~ a_{[1]}~+~(n-1)\cdot d \]

   Find \(a_{[41]}\): the value of the element with an index equaling 41.

   
   

   Solution

   Show solution

   Use the explicit formula.

   \[a_{[41]} = 22+(41-1)\cdot 3 \] \[a_{[41]} = 142 \]

   
   
4. 
   

   Question

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   Below, a specific arithmetic sequence, with a first element equal 11 and common difference equal 14, is defined recursively, where \(n\) is any whole number larger than 1. \[a_{[1]} = 11\] \[d = 14\] \[a_{[n]} = a_{[n-1]}+d\] Using that definition we can find the first few terms. \[a_{[2]} ~=~ a_{[1]}+d ~=~ 11+14 ~=~ 25 \] \[a_{[3]} ~=~ a_{[2]}+d ~=~ 25+14 ~=~ 39 \] \[a_{[4]} ~=~ a_{[3]}+d ~=~ 39+14 ~=~ 53 \] \[a_{[5]} ~=~ a_{[4]}+d ~=~ 53+14 ~=~ 67 \]

   We say the first element is 11, the second element is 25, the third element is 39, the fourth element is 53, and the fifth element is 67.

   The index is the address, so \(a_{[2]}=25\) says that the term with an index of 2 has a value of 25.

   To find the value of an element with a large index, it is useful to know the explicit formula:

   \[a_{[n]} ~~=~~ a_{[1]}~+~(n-1)\cdot d \]

   Find \(a_{[77]}\): the value of the element with an index equaling 77.

   
   

   Solution

   Show solution

   Use the explicit formula.

   \[a_{[77]} = 11+(77-1)\cdot 14 \] \[a_{[77]} = 1075 \]

   
   
5. 
   

   Question

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   Below, a specific arithmetic sequence, with a first element equal 2 and common difference equal 7, is defined recursively, where \(n\) is any whole number larger than 1. \[a_{[1]} = 2\] \[d = 7\] \[a_{[n]} = a_{[n-1]}+d\] Using that definition we can find the first few terms. \[a_{[2]} ~=~ a_{[1]}+d ~=~ 2+7 ~=~ 9 \] \[a_{[3]} ~=~ a_{[2]}+d ~=~ 9+7 ~=~ 16 \] \[a_{[4]} ~=~ a_{[3]}+d ~=~ 16+7 ~=~ 23 \] \[a_{[5]} ~=~ a_{[4]}+d ~=~ 23+7 ~=~ 30 \]

   We say the first element is 2, the second element is 9, the third element is 16, the fourth element is 23, and the fifth element is 30.

   The index is the address, so \(a_{[2]}=9\) says that the term with an index of 2 has a value of 9.

   To find the value of an element with a large index, it is useful to know the explicit formula:

   \[a_{[n]} ~~=~~ a_{[1]}~+~(n-1)\cdot d \]

   Find \(a_{[47]}\): the value of the element with an index equaling 47.

   
   

   Solution

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   Use the explicit formula.

   \[a_{[47]} = 2+(47-1)\cdot 7 \] \[a_{[47]} = 324 \]

   
   
6. 
   

   Question

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   Below, a specific arithmetic sequence, with a first element equal 20 and common difference equal 14, is defined recursively, where \(n\) is any whole number larger than 1. \[a_{[1]} = 20\] \[d = 14\] \[a_{[n]} = a_{[n-1]}+d\] Using that definition we can find the first few terms. \[a_{[2]} ~=~ a_{[1]}+d ~=~ 20+14 ~=~ 34 \] \[a_{[3]} ~=~ a_{[2]}+d ~=~ 34+14 ~=~ 48 \] \[a_{[4]} ~=~ a_{[3]}+d ~=~ 48+14 ~=~ 62 \] \[a_{[5]} ~=~ a_{[4]}+d ~=~ 62+14 ~=~ 76 \]

   We say the first element is 20, the second element is 34, the third element is 48, the fourth element is 62, and the fifth element is 76.

   The index is the address, so \(a_{[2]}=34\) says that the term with an index of 2 has a value of 34.

   To find the value of an element with a large index, it is useful to know the explicit formula:

   \[a_{[n]} ~~=~~ a_{[1]}~+~(n-1)\cdot d \]

   Find \(a_{[69]}\): the value of the element with an index equaling 69.

   
   

   Solution

   Show solution

   Use the explicit formula.

   \[a_{[69]} = 20+(69-1)\cdot 14 \] \[a_{[69]} = 972 \]

   
   
7. 
   

   Question

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   Below, a specific arithmetic sequence, with a first element equal 5 and common difference equal 9, is defined recursively, where \(n\) is any whole number larger than 1. \[a_{[1]} = 5\] \[d = 9\] \[a_{[n]} = a_{[n-1]}+d\] Using that definition we can find the first few terms. \[a_{[2]} ~=~ a_{[1]}+d ~=~ 5+9 ~=~ 14 \] \[a_{[3]} ~=~ a_{[2]}+d ~=~ 14+9 ~=~ 23 \] \[a_{[4]} ~=~ a_{[3]}+d ~=~ 23+9 ~=~ 32 \] \[a_{[5]} ~=~ a_{[4]}+d ~=~ 32+9 ~=~ 41 \]

   We say the first element is 5, the second element is 14, the third element is 23, the fourth element is 32, and the fifth element is 41.

   The index is the address, so \(a_{[2]}=14\) says that the term with an index of 2 has a value of 14.

   To find the value of an element with a large index, it is useful to know the explicit formula:

   \[a_{[n]} ~~=~~ a_{[1]}~+~(n-1)\cdot d \]

   Find \(a_{[51]}\): the value of the element with an index equaling 51.

   
   

   Solution

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   Use the explicit formula.

   \[a_{[51]} = 5+(51-1)\cdot 9 \] \[a_{[51]} = 455 \]

   
   
8. 
   

   Question

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   Below, a specific arithmetic sequence, with a first element equal 33 and common difference equal 9, is defined recursively, where \(n\) is any whole number larger than 1. \[a_{[1]} = 33\] \[d = 9\] \[a_{[n]} = a_{[n-1]}+d\] Using that definition we can find the first few terms. \[a_{[2]} ~=~ a_{[1]}+d ~=~ 33+9 ~=~ 42 \] \[a_{[3]} ~=~ a_{[2]}+d ~=~ 42+9 ~=~ 51 \] \[a_{[4]} ~=~ a_{[3]}+d ~=~ 51+9 ~=~ 60 \] \[a_{[5]} ~=~ a_{[4]}+d ~=~ 60+9 ~=~ 69 \]

   We say the first element is 33, the second element is 42, the third element is 51, the fourth element is 60, and the fifth element is 69.

   The index is the address, so \(a_{[2]}=42\) says that the term with an index of 2 has a value of 42.

   To find the value of an element with a large index, it is useful to know the explicit formula:

   \[a_{[n]} ~~=~~ a_{[1]}~+~(n-1)\cdot d \]

   Find \(a_{[65]}\): the value of the element with an index equaling 65.

   
   

   Solution

   Show solution

   Use the explicit formula.

   \[a_{[65]} = 33+(65-1)\cdot 9 \] \[a_{[65]} = 609 \]

   
   
9. 
   

   Question

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   Below, a specific arithmetic sequence, with a first element equal 14 and common difference equal 3, is defined recursively, where \(n\) is any whole number larger than 1. \[a_{[1]} = 14\] \[d = 3\] \[a_{[n]} = a_{[n-1]}+d\] Using that definition we can find the first few terms. \[a_{[2]} ~=~ a_{[1]}+d ~=~ 14+3 ~=~ 17 \] \[a_{[3]} ~=~ a_{[2]}+d ~=~ 17+3 ~=~ 20 \] \[a_{[4]} ~=~ a_{[3]}+d ~=~ 20+3 ~=~ 23 \] \[a_{[5]} ~=~ a_{[4]}+d ~=~ 23+3 ~=~ 26 \]

   We say the first element is 14, the second element is 17, the third element is 20, the fourth element is 23, and the fifth element is 26.

   The index is the address, so \(a_{[2]}=17\) says that the term with an index of 2 has a value of 17.

   To find the value of an element with a large index, it is useful to know the explicit formula:

   \[a_{[n]} ~~=~~ a_{[1]}~+~(n-1)\cdot d \]

   Find \(a_{[77]}\): the value of the element with an index equaling 77.

   
   

   Solution

   Show solution

   Use the explicit formula.

   \[a_{[77]} = 14+(77-1)\cdot 3 \] \[a_{[77]} = 242 \]

   
   
10. 
    

    Question

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    Below, a specific arithmetic sequence, with a first element equal 17 and common difference equal 9, is defined recursively, where \(n\) is any whole number larger than 1. \[a_{[1]} = 17\] \[d = 9\] \[a_{[n]} = a_{[n-1]}+d\] Using that definition we can find the first few terms. \[a_{[2]} ~=~ a_{[1]}+d ~=~ 17+9 ~=~ 26 \] \[a_{[3]} ~=~ a_{[2]}+d ~=~ 26+9 ~=~ 35 \] \[a_{[4]} ~=~ a_{[3]}+d ~=~ 35+9 ~=~ 44 \] \[a_{[5]} ~=~ a_{[4]}+d ~=~ 44+9 ~=~ 53 \]

    We say the first element is 17, the second element is 26, the third element is 35, the fourth element is 44, and the fifth element is 53.

    The index is the address, so \(a_{[2]}=26\) says that the term with an index of 2 has a value of 26.

    To find the value of an element with a large index, it is useful to know the explicit formula:

    \[a_{[n]} ~~=~~ a_{[1]}~+~(n-1)\cdot d \]

    Find \(a_{[79]}\): the value of the element with an index equaling 79.

    
    

    Solution

    Show solution

    Use the explicit formula.

    \[a_{[79]} = 17+(79-1)\cdot 9 \] \[a_{[79]} = 719 \]

    
    
11. 
    

    Question

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    Below, a specific arithmetic sequence, with a first element equal 6 and common difference equal 13, is defined recursively, where \(n\) is any whole number larger than 1. \[a_{[1]} = 6\] \[a_{[n]} = a_{[n-1]}+13\]

    Find the index of the element with a value of 370. In other words, find \(i\) such that \(a_{[i]}=370\).

    
    

    Solution

    Show solution

    Start with the explicit formula.

    \[a_n = a_1 + (n-1)d \]

    \[370 = 6+(i-1)\cdot 13 \]

    Subtract 6.

    \[364 = (i-1)\cdot 13 \]

    Divide by 13.

    \[28 = i-1 \]

    Add 1.

    \[29 = i \]

    Double check.

    \[a_{[29]} = 6+(29-1)\cdot 13 = 370 \]

    If you are fancy, you can rearrange the explicit formula to create a new formula.

    \[a_n = a_1+(n-1)d \]

    \[a_n-a_1 = (n-1)d \]

    \[\frac{a_n-a_1}{d} = n-1 \]

    \[n = \frac{a_n-a_1}{d}+1 \]

    
    
12. 
    

    Question

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    Below, a specific arithmetic sequence, with a first element equal 12 and common difference equal 13, is defined recursively, where \(n\) is any whole number larger than 1. \[a_{[1]} = 12\] \[a_{[n]} = a_{[n-1]}+13\]

    Find the index of the element with a value of 415. In other words, find \(i\) such that \(a_{[i]}=415\).

    
    

    Solution

    Show solution

    Start with the explicit formula.

    \[a_n = a_1 + (n-1)d \]

    \[415 = 12+(i-1)\cdot 13 \]

    Subtract 12.

    \[403 = (i-1)\cdot 13 \]

    Divide by 13.

    \[31 = i-1 \]

    Add 1.

    \[32 = i \]

    Double check.

    \[a_{[32]} = 12+(32-1)\cdot 13 = 415 \]

    If you are fancy, you can rearrange the explicit formula to create a new formula.

    \[a_n = a_1+(n-1)d \]

    \[a_n-a_1 = (n-1)d \]

    \[\frac{a_n-a_1}{d} = n-1 \]

    \[n = \frac{a_n-a_1}{d}+1 \]

    
    
13. 
    

    Question

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    Below, a specific arithmetic sequence, with a first element equal 48 and common difference equal 12, is defined recursively, where \(n\) is any whole number larger than 1. \[a_{[1]} = 48\] \[a_{[n]} = a_{[n-1]}+12\]

    Find the index of the element with a value of 324. In other words, find \(i\) such that \(a_{[i]}=324\).

    
    

    Solution

    Show solution

    Start with the explicit formula.

    \[a_n = a_1 + (n-1)d \]

    \[324 = 48+(i-1)\cdot 12 \]

    Subtract 48.

    \[276 = (i-1)\cdot 12 \]

    Divide by 12.

    \[23 = i-1 \]

    Add 1.

    \[24 = i \]

    Double check.

    \[a_{[24]} = 48+(24-1)\cdot 12 = 324 \]

    If you are fancy, you can rearrange the explicit formula to create a new formula.

    \[a_n = a_1+(n-1)d \]

    \[a_n-a_1 = (n-1)d \]

    \[\frac{a_n-a_1}{d} = n-1 \]

    \[n = \frac{a_n-a_1}{d}+1 \]

    
    
14. 
    

    Question

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    Below, a specific arithmetic sequence, with a first element equal 39 and common difference equal 5, is defined recursively, where \(n\) is any whole number larger than 1. \[a_{[1]} = 39\] \[a_{[n]} = a_{[n-1]}+5\]

    Find the index of the element with a value of 169. In other words, find \(i\) such that \(a_{[i]}=169\).

    
    

    Solution

    Show solution

    Start with the explicit formula.

    \[a_n = a_1 + (n-1)d \]

    \[169 = 39+(i-1)\cdot 5 \]

    Subtract 39.

    \[130 = (i-1)\cdot 5 \]

    Divide by 5.

    \[26 = i-1 \]

    Add 1.

    \[27 = i \]

    Double check.

    \[a_{[27]} = 39+(27-1)\cdot 5 = 169 \]

    If you are fancy, you can rearrange the explicit formula to create a new formula.

    \[a_n = a_1+(n-1)d \]

    \[a_n-a_1 = (n-1)d \]

    \[\frac{a_n-a_1}{d} = n-1 \]

    \[n = \frac{a_n-a_1}{d}+1 \]

    
    
15. 
    

    Question

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    Below, a specific arithmetic sequence, with a first element equal 29 and common difference equal 4, is defined recursively, where \(n\) is any whole number larger than 1. \[a_{[1]} = 29\] \[a_{[n]} = a_{[n-1]}+4\]

    Find the index of the element with a value of 141. In other words, find \(i\) such that \(a_{[i]}=141\).

    
    

    Solution

    Show solution

    Start with the explicit formula.

    \[a_n = a_1 + (n-1)d \]

    \[141 = 29+(i-1)\cdot 4 \]

    Subtract 29.

    \[112 = (i-1)\cdot 4 \]

    Divide by 4.

    \[28 = i-1 \]

    Add 1.

    \[29 = i \]

    Double check.

    \[a_{[29]} = 29+(29-1)\cdot 4 = 141 \]

    If you are fancy, you can rearrange the explicit formula to create a new formula.

    \[a_n = a_1+(n-1)d \]

    \[a_n-a_1 = (n-1)d \]

    \[\frac{a_n-a_1}{d} = n-1 \]

    \[n = \frac{a_n-a_1}{d}+1 \]

    
    
16. 
    

    Question

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    Below, a specific arithmetic sequence, with a first element equal 7 and common difference equal 10, is defined recursively, where \(n\) is any whole number larger than 1. \[a_{[1]} = 7\] \[a_{[n]} = a_{[n-1]}+10\]

    Find the index of the element with a value of 307. In other words, find \(i\) such that \(a_{[i]}=307\).

    
    

    Solution

    Show solution

    Start with the explicit formula.

    \[a_n = a_1 + (n-1)d \]

    \[307 = 7+(i-1)\cdot 10 \]

    Subtract 7.

    \[300 = (i-1)\cdot 10 \]

    Divide by 10.

    \[30 = i-1 \]

    Add 1.

    \[31 = i \]

    Double check.

    \[a_{[31]} = 7+(31-1)\cdot 10 = 307 \]

    If you are fancy, you can rearrange the explicit formula to create a new formula.

    \[a_n = a_1+(n-1)d \]

    \[a_n-a_1 = (n-1)d \]

    \[\frac{a_n-a_1}{d} = n-1 \]

    \[n = \frac{a_n-a_1}{d}+1 \]

    
    
17. 
    

    Question

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    Below, a specific arithmetic sequence, with a first element equal 37 and common difference equal 3, is defined recursively, where \(n\) is any whole number larger than 1. \[a_{[1]} = 37\] \[a_{[n]} = a_{[n-1]}+3\]

    Find the index of the element with a value of 109. In other words, find \(i\) such that \(a_{[i]}=109\).

    
    

    Solution

    Show solution

    Start with the explicit formula.

    \[a_n = a_1 + (n-1)d \]

    \[109 = 37+(i-1)\cdot 3 \]

    Subtract 37.

    \[72 = (i-1)\cdot 3 \]

    Divide by 3.

    \[24 = i-1 \]

    Add 1.

    \[25 = i \]

    Double check.

    \[a_{[25]} = 37+(25-1)\cdot 3 = 109 \]

    If you are fancy, you can rearrange the explicit formula to create a new formula.

    \[a_n = a_1+(n-1)d \]

    \[a_n-a_1 = (n-1)d \]

    \[\frac{a_n-a_1}{d} = n-1 \]

    \[n = \frac{a_n-a_1}{d}+1 \]

    
    
18. 
    

    Question

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    Below, a specific arithmetic sequence, with a first element equal 22 and common difference equal 4, is defined recursively, where \(n\) is any whole number larger than 1. \[a_{[1]} = 22\] \[a_{[n]} = a_{[n-1]}+4\]

    Find the index of the element with a value of 138. In other words, find \(i\) such that \(a_{[i]}=138\).

    
    

    Solution

    Show solution

    Start with the explicit formula.

    \[a_n = a_1 + (n-1)d \]

    \[138 = 22+(i-1)\cdot 4 \]

    Subtract 22.

    \[116 = (i-1)\cdot 4 \]

    Divide by 4.

    \[29 = i-1 \]

    Add 1.

    \[30 = i \]

    Double check.

    \[a_{[30]} = 22+(30-1)\cdot 4 = 138 \]

    If you are fancy, you can rearrange the explicit formula to create a new formula.

    \[a_n = a_1+(n-1)d \]

    \[a_n-a_1 = (n-1)d \]

    \[\frac{a_n-a_1}{d} = n-1 \]

    \[n = \frac{a_n-a_1}{d}+1 \]

    
    
19. 
    

    Question

    prevnext

    Below, a specific arithmetic sequence, with a first element equal 10 and common difference equal 11, is defined recursively, where \(n\) is any whole number larger than 1. \[a_{[1]} = 10\] \[a_{[n]} = a_{[n-1]}+11\]

    Find the index of the element with a value of 395. In other words, find \(i\) such that \(a_{[i]}=395\).

    
    

    Solution

    Show solution

    Start with the explicit formula.

    \[a_n = a_1 + (n-1)d \]

    \[395 = 10+(i-1)\cdot 11 \]

    Subtract 10.

    \[385 = (i-1)\cdot 11 \]

    Divide by 11.

    \[35 = i-1 \]

    Add 1.

    \[36 = i \]

    Double check.

    \[a_{[36]} = 10+(36-1)\cdot 11 = 395 \]

    If you are fancy, you can rearrange the explicit formula to create a new formula.

    \[a_n = a_1+(n-1)d \]

    \[a_n-a_1 = (n-1)d \]

    \[\frac{a_n-a_1}{d} = n-1 \]

    \[n = \frac{a_n-a_1}{d}+1 \]

    
    
20. 
    

    Question

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    Below, a specific arithmetic sequence, with a first element equal 22 and common difference equal 12, is defined recursively, where \(n\) is any whole number larger than 1. \[a_{[1]} = 22\] \[a_{[n]} = a_{[n-1]}+12\]

    Find the index of the element with a value of 310. In other words, find \(i\) such that \(a_{[i]}=310\).

    
    

    Solution

    Show solution

    Start with the explicit formula.

    \[a_n = a_1 + (n-1)d \]

    \[310 = 22+(i-1)\cdot 12 \]

    Subtract 22.

    \[288 = (i-1)\cdot 12 \]

    Divide by 12.

    \[24 = i-1 \]

    Add 1.

    \[25 = i \]

    Double check.

    \[a_{[25]} = 22+(25-1)\cdot 12 = 310 \]

    If you are fancy, you can rearrange the explicit formula to create a new formula.

    \[a_n = a_1+(n-1)d \]

    \[a_n-a_1 = (n-1)d \]

    \[\frac{a_n-a_1}{d} = n-1 \]

    \[n = \frac{a_n-a_1}{d}+1 \]

    
    
21. 
    

    Question

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    Let \(a_{i}\) represent the \(i\)th element of an arithmetic sequence, and \[a_{13} = 180\] \[a_{32} = 408\]

    Find the common difference.

    
    

    Solution

    Show solution

    Start with the explicit formula.

    \[a_n = a_1 + (n-1)d \]

    Fill in the knowns, resulting in a system of 2 equations and 2 variables.

    \[180 = a_1+(13-1)d \] \[408 = a_1+(32-1)d \]

    Simplify the parentheses.

    \[180 = a_1+12d \] \[408 = a_1+31d \]

    Subtract the first equation from the second to eliminate \(a_1\).

    \[228 = 19d \]

    Divide by 19.

    \[12 = d \]

    
    
22. 
    

    Question

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    Let \(a_{i}\) represent the \(i\)th element of an arithmetic sequence, and \[a_{30} = 433\] \[a_{48} = 685\]

    Find the common difference.

    
    

    Solution

    Show solution

    Start with the explicit formula.

    \[a_n = a_1 + (n-1)d \]

    Fill in the knowns, resulting in a system of 2 equations and 2 variables.

    \[433 = a_1+(30-1)d \] \[685 = a_1+(48-1)d \]

    Simplify the parentheses.

    \[433 = a_1+29d \] \[685 = a_1+47d \]

    Subtract the first equation from the second to eliminate \(a_1\).

    \[252 = 18d \]

    Divide by 18.

    \[14 = d \]

    
    
23. 
    

    Question

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    Let \(a_{i}\) represent the \(i\)th element of an arithmetic sequence, and \[a_{27} = 298\] \[a_{43} = 458\]

    Find the common difference.

    
    

    Solution

    Show solution

    Start with the explicit formula.

    \[a_n = a_1 + (n-1)d \]

    Fill in the knowns, resulting in a system of 2 equations and 2 variables.

    \[298 = a_1+(27-1)d \] \[458 = a_1+(43-1)d \]

    Simplify the parentheses.

    \[298 = a_1+26d \] \[458 = a_1+42d \]

    Subtract the first equation from the second to eliminate \(a_1\).

    \[160 = 16d \]

    Divide by 16.

    \[10 = d \]

    
    
24. 
    

    Question

    prevnext

    Let \(a_{i}\) represent the \(i\)th element of an arithmetic sequence, and \[a_{19} = 123\] \[a_{34} = 198\]

    Find the common difference.

    
    

    Solution

    Show solution

    Start with the explicit formula.

    \[a_n = a_1 + (n-1)d \]

    Fill in the knowns, resulting in a system of 2 equations and 2 variables.

    \[123 = a_1+(19-1)d \] \[198 = a_1+(34-1)d \]

    Simplify the parentheses.

    \[123 = a_1+18d \] \[198 = a_1+33d \]

    Subtract the first equation from the second to eliminate \(a_1\).

    \[75 = 15d \]

    Divide by 15.

    \[5 = d \]

    
    
25. 
    

    Question

    prevnext

    Let \(a_{i}\) represent the \(i\)th element of an arithmetic sequence, and \[a_{11} = 140\] \[a_{20} = 257\]

    Find the common difference.

    
    

    Solution

    Show solution

    Start with the explicit formula.

    \[a_n = a_1 + (n-1)d \]

    Fill in the knowns, resulting in a system of 2 equations and 2 variables.

    \[140 = a_1+(11-1)d \] \[257 = a_1+(20-1)d \]

    Simplify the parentheses.

    \[140 = a_1+10d \] \[257 = a_1+19d \]

    Subtract the first equation from the second to eliminate \(a_1\).

    \[117 = 9d \]

    Divide by 9.

    \[13 = d \]

    
    
26. 
    

    Question

    prevnext

    Let \(a_{i}\) represent the \(i\)th element of an arithmetic sequence, and \[a_{35} = 268\] \[a_{53} = 394\]

    Find the common difference.

    
    

    Solution

    Show solution

    Start with the explicit formula.

    \[a_n = a_1 + (n-1)d \]

    Fill in the knowns, resulting in a system of 2 equations and 2 variables.

    \[268 = a_1+(35-1)d \] \[394 = a_1+(53-1)d \]

    Simplify the parentheses.

    \[268 = a_1+34d \] \[394 = a_1+52d \]

    Subtract the first equation from the second to eliminate \(a_1\).

    \[126 = 18d \]

    Divide by 18.

    \[7 = d \]

    
    
27. 
    

    Question

    prevnext

    Let \(a_{i}\) represent the \(i\)th element of an arithmetic sequence, and \[a_{24} = 140\] \[a_{36} = 188\]

    Find the common difference.

    
    

    Solution

    Show solution

    Start with the explicit formula.

    \[a_n = a_1 + (n-1)d \]

    Fill in the knowns, resulting in a system of 2 equations and 2 variables.

    \[140 = a_1+(24-1)d \] \[188 = a_1+(36-1)d \]

    Simplify the parentheses.

    \[140 = a_1+23d \] \[188 = a_1+35d \]

    Subtract the first equation from the second to eliminate \(a_1\).

    \[48 = 12d \]

    Divide by 12.

    \[4 = d \]

    
    
28. 
    

    Question

    prevnext

    Let \(a_{i}\) represent the \(i\)th element of an arithmetic sequence, and \[a_{40} = 531\] \[a_{50} = 661\]

    Find the common difference.

    
    

    Solution

    Show solution

    Start with the explicit formula.

    \[a_n = a_1 + (n-1)d \]

    Fill in the knowns, resulting in a system of 2 equations and 2 variables.

    \[531 = a_1+(40-1)d \] \[661 = a_1+(50-1)d \]

    Simplify the parentheses.

    \[531 = a_1+39d \] \[661 = a_1+49d \]

    Subtract the first equation from the second to eliminate \(a_1\).

    \[130 = 10d \]

    Divide by 10.

    \[13 = d \]

    
    
29. 
    

    Question

    prevnext

    Let \(a_{i}\) represent the \(i\)th element of an arithmetic sequence, and \[a_{14} = 83\] \[a_{33} = 178\]

    Find the common difference.

    
    

    Solution

    Show solution

    Start with the explicit formula.

    \[a_n = a_1 + (n-1)d \]

    Fill in the knowns, resulting in a system of 2 equations and 2 variables.

    \[83 = a_1+(14-1)d \] \[178 = a_1+(33-1)d \]

    Simplify the parentheses.

    \[83 = a_1+13d \] \[178 = a_1+32d \]

    Subtract the first equation from the second to eliminate \(a_1\).

    \[95 = 19d \]

    Divide by 19.

    \[5 = d \]

    
    
30. 
    

    Question

    prevnext

    Let \(a_{i}\) represent the \(i\)th element of an arithmetic sequence, and \[a_{32} = 514\] \[a_{40} = 634\]

    Find the common difference.

    
    

    Solution

    Show solution

    Start with the explicit formula.

    \[a_n = a_1 + (n-1)d \]

    Fill in the knowns, resulting in a system of 2 equations and 2 variables.

    \[514 = a_1+(32-1)d \] \[634 = a_1+(40-1)d \]

    Simplify the parentheses.

    \[514 = a_1+31d \] \[634 = a_1+39d \]

    Subtract the first equation from the second to eliminate \(a_1\).

    \[120 = 8d \]

    Divide by 8.

    \[15 = d \]

    
    
31. 
    

    Question

    prevnext

    Let \(a_{i}\) represent the \(i\)th term of an arithmetic series, and \[a_1=9\] \[a_2=15\]

    Find the sum of the first 16 terms. In other words, \[\sum_{i=1}^{16}a_i = \,?\]

    
    

    Solution

    Show solution

    The sum of an arithmetic series is found by averaging the first and last terms and multiplying by the number of terms.

    \[\sum_{i=p}^q a_i ~~=~~ \frac{a_p+a_q}{2} \cdot (q-p+1)\]

    Determine the common difference.

    \[d ~=~ (15)-(9) ~=~ 6\]

    Find the value of \(a_{16}\) using the explicit arithmetic formula.

    \[a_{n} = a_1+(n-1)d \] \[a_{16} = 9+(16-1)(6)\] \[a_{16} = 99\]

    So,

    \[\sum_{i=1}^{16} a_i ~=~ \frac{(9)+(99)}{2} \cdot (16)\]

    Evaluate. \[\sum_{i=1}^{16} a_i ~=~ 864\]

    I think of an arithmetic series as representing stacks of coins in a stair pattern. Our goal is to count the coins.

    
    

    Because of the regularity of the step size, the average stack height is simply the average of the first and last stacks.

    If we redistribute the wealth, then all the stacks have the average height.

    
    

    Thus, we just need to multiply the number of stacks by the average height. And the average height is simply the average of the first and last stack. \(54\cdot16 = 864\).

    
    
32. 
    

    Question

    prevnext

    Let \(a_{i}\) represent the \(i\)th term of an arithmetic series, and \[a_1=11\] \[a_2=13\]

    Find the sum of the first 39 terms. In other words, \[\sum_{i=1}^{39}a_i = \,?\]

    
    

    Solution

    Show solution

    The sum of an arithmetic series is found by averaging the first and last terms and multiplying by the number of terms.

    \[\sum_{i=p}^q a_i ~~=~~ \frac{a_p+a_q}{2} \cdot (q-p+1)\]

    Determine the common difference.

    \[d ~=~ (13)-(11) ~=~ 2\]

    Find the value of \(a_{39}\) using the explicit arithmetic formula.

    \[a_{n} = a_1+(n-1)d \] \[a_{39} = 11+(39-1)(2)\] \[a_{39} = 87\]

    So,

    \[\sum_{i=1}^{39} a_i ~=~ \frac{(11)+(87)}{2} \cdot (39)\]

    Evaluate. \[\sum_{i=1}^{39} a_i ~=~ 1911\]

    I think of an arithmetic series as representing stacks of coins in a stair pattern. Our goal is to count the coins.

    
    

    Because of the regularity of the step size, the average stack height is simply the average of the first and last stacks.

    If we redistribute the wealth, then all the stacks have the average height.

    
    

    Thus, we just need to multiply the number of stacks by the average height. And the average height is simply the average of the first and last stack. \(49\cdot39 = 1911\).

    
    
33. 
    

    Question

    prevnext

    Let \(a_{i}\) represent the \(i\)th term of an arithmetic series, and \[a_1=20\] \[a_2=24\]

    Find the sum of the first 35 terms. In other words, \[\sum_{i=1}^{35}a_i = \,?\]

    
    

    Solution

    Show solution

    The sum of an arithmetic series is found by averaging the first and last terms and multiplying by the number of terms.

    \[\sum_{i=p}^q a_i ~~=~~ \frac{a_p+a_q}{2} \cdot (q-p+1)\]

    Determine the common difference.

    \[d ~=~ (24)-(20) ~=~ 4\]

    Find the value of \(a_{35}\) using the explicit arithmetic formula.

    \[a_{n} = a_1+(n-1)d \] \[a_{35} = 20+(35-1)(4)\] \[a_{35} = 156\]

    So,

    \[\sum_{i=1}^{35} a_i ~=~ \frac{(20)+(156)}{2} \cdot (35)\]

    Evaluate. \[\sum_{i=1}^{35} a_i ~=~ 3080\]

    I think of an arithmetic series as representing stacks of coins in a stair pattern. Our goal is to count the coins.

    
    

    Because of the regularity of the step size, the average stack height is simply the average of the first and last stacks.

    If we redistribute the wealth, then all the stacks have the average height.

    
    

    Thus, we just need to multiply the number of stacks by the average height. And the average height is simply the average of the first and last stack. \(88\cdot35 = 3080\).

    
    
34. 
    

    Question

    prevnext

    Let \(a_{i}\) represent the \(i\)th term of an arithmetic series, and \[a_1=12\] \[a_2=15\]

    Find the sum of the first 15 terms. In other words, \[\sum_{i=1}^{15}a_i = \,?\]

    
    

    Solution

    Show solution

    The sum of an arithmetic series is found by averaging the first and last terms and multiplying by the number of terms.

    \[\sum_{i=p}^q a_i ~~=~~ \frac{a_p+a_q}{2} \cdot (q-p+1)\]

    Determine the common difference.

    \[d ~=~ (15)-(12) ~=~ 3\]

    Find the value of \(a_{15}\) using the explicit arithmetic formula.

    \[a_{n} = a_1+(n-1)d \] \[a_{15} = 12+(15-1)(3)\] \[a_{15} = 54\]

    So,

    \[\sum_{i=1}^{15} a_i ~=~ \frac{(12)+(54)}{2} \cdot (15)\]

    Evaluate. \[\sum_{i=1}^{15} a_i ~=~ 495\]

    I think of an arithmetic series as representing stacks of coins in a stair pattern. Our goal is to count the coins.

    
    

    Because of the regularity of the step size, the average stack height is simply the average of the first and last stacks.

    If we redistribute the wealth, then all the stacks have the average height.

    
    

    Thus, we just need to multiply the number of stacks by the average height. And the average height is simply the average of the first and last stack. \(33\cdot15 = 495\).

    
    
35. 
    

    Question

    prevnext

    Let \(a_{i}\) represent the \(i\)th term of an arithmetic series, and \[a_1=6\] \[a_2=9\]

    Find the sum of the first 17 terms. In other words, \[\sum_{i=1}^{17}a_i = \,?\]

    
    

    Solution

    Show solution

    The sum of an arithmetic series is found by averaging the first and last terms and multiplying by the number of terms.

    \[\sum_{i=p}^q a_i ~~=~~ \frac{a_p+a_q}{2} \cdot (q-p+1)\]

    Determine the common difference.

    \[d ~=~ (9)-(6) ~=~ 3\]

    Find the value of \(a_{17}\) using the explicit arithmetic formula.

    \[a_{n} = a_1+(n-1)d \] \[a_{17} = 6+(17-1)(3)\] \[a_{17} = 54\]

    So,

    \[\sum_{i=1}^{17} a_i ~=~ \frac{(6)+(54)}{2} \cdot (17)\]

    Evaluate. \[\sum_{i=1}^{17} a_i ~=~ 510\]

    I think of an arithmetic series as representing stacks of coins in a stair pattern. Our goal is to count the coins.

    
    

    Because of the regularity of the step size, the average stack height is simply the average of the first and last stacks.

    If we redistribute the wealth, then all the stacks have the average height.

    
    

    Thus, we just need to multiply the number of stacks by the average height. And the average height is simply the average of the first and last stack. \(30\cdot17 = 510\).

    
    
36. 
    

    Question

    prevnext

    Let \(a_{i}\) represent the \(i\)th term of an arithmetic series, and \[a_1=15\] \[a_2=21\]

    Find the sum of the first 32 terms. In other words, \[\sum_{i=1}^{32}a_i = \,?\]

    
    

    Solution

    Show solution

    The sum of an arithmetic series is found by averaging the first and last terms and multiplying by the number of terms.

    \[\sum_{i=p}^q a_i ~~=~~ \frac{a_p+a_q}{2} \cdot (q-p+1)\]

    Determine the common difference.

    \[d ~=~ (21)-(15) ~=~ 6\]

    Find the value of \(a_{32}\) using the explicit arithmetic formula.

    \[a_{n} = a_1+(n-1)d \] \[a_{32} = 15+(32-1)(6)\] \[a_{32} = 201\]

    So,

    \[\sum_{i=1}^{32} a_i ~=~ \frac{(15)+(201)}{2} \cdot (32)\]

    Evaluate. \[\sum_{i=1}^{32} a_i ~=~ 3456\]

    I think of an arithmetic series as representing stacks of coins in a stair pattern. Our goal is to count the coins.

    
    

    Because of the regularity of the step size, the average stack height is simply the average of the first and last stacks.

    If we redistribute the wealth, then all the stacks have the average height.

    
    

    Thus, we just need to multiply the number of stacks by the average height. And the average height is simply the average of the first and last stack. \(108\cdot32 = 3456\).

    
    
37. 
    

    Question

    prevnext

    Let \(a_{i}\) represent the \(i\)th term of an arithmetic series, and \[a_1=17\] \[a_2=19\]

    Find the sum of the first 16 terms. In other words, \[\sum_{i=1}^{16}a_i = \,?\]

    
    

    Solution

    Show solution

    The sum of an arithmetic series is found by averaging the first and last terms and multiplying by the number of terms.

    \[\sum_{i=p}^q a_i ~~=~~ \frac{a_p+a_q}{2} \cdot (q-p+1)\]

    Determine the common difference.

    \[d ~=~ (19)-(17) ~=~ 2\]

    Find the value of \(a_{16}\) using the explicit arithmetic formula.

    \[a_{n} = a_1+(n-1)d \] \[a_{16} = 17+(16-1)(2)\] \[a_{16} = 47\]

    So,

    \[\sum_{i=1}^{16} a_i ~=~ \frac{(17)+(47)}{2} \cdot (16)\]

    Evaluate. \[\sum_{i=1}^{16} a_i ~=~ 512\]

    I think of an arithmetic series as representing stacks of coins in a stair pattern. Our goal is to count the coins.

    
    

    Because of the regularity of the step size, the average stack height is simply the average of the first and last stacks.

    If we redistribute the wealth, then all the stacks have the average height.

    
    

    Thus, we just need to multiply the number of stacks by the average height. And the average height is simply the average of the first and last stack. \(32\cdot16 = 512\).

    
    
38. 
    

    Question

    prevnext

    Let \(a_{i}\) represent the \(i\)th term of an arithmetic series, and \[a_1=13\] \[a_2=15\]

    Find the sum of the first 40 terms. In other words, \[\sum_{i=1}^{40}a_i = \,?\]

    
    

    Solution

    Show solution

    The sum of an arithmetic series is found by averaging the first and last terms and multiplying by the number of terms.

    \[\sum_{i=p}^q a_i ~~=~~ \frac{a_p+a_q}{2} \cdot (q-p+1)\]

    Determine the common difference.

    \[d ~=~ (15)-(13) ~=~ 2\]

    Find the value of \(a_{40}\) using the explicit arithmetic formula.

    \[a_{n} = a_1+(n-1)d \] \[a_{40} = 13+(40-1)(2)\] \[a_{40} = 91\]

    So,

    \[\sum_{i=1}^{40} a_i ~=~ \frac{(13)+(91)}{2} \cdot (40)\]

    Evaluate. \[\sum_{i=1}^{40} a_i ~=~ 2080\]

    I think of an arithmetic series as representing stacks of coins in a stair pattern. Our goal is to count the coins.

    
    

    Because of the regularity of the step size, the average stack height is simply the average of the first and last stacks.

    If we redistribute the wealth, then all the stacks have the average height.

    
    

    Thus, we just need to multiply the number of stacks by the average height. And the average height is simply the average of the first and last stack. \(52\cdot40 = 2080\).

    
    
39. 
    

    Question

    prevnext

    Let \(a_{i}\) represent the \(i\)th term of an arithmetic series, and \[a_1=17\] \[a_2=20\]

    Find the sum of the first 15 terms. In other words, \[\sum_{i=1}^{15}a_i = \,?\]

    
    

    Solution

    Show solution

    The sum of an arithmetic series is found by averaging the first and last terms and multiplying by the number of terms.

    \[\sum_{i=p}^q a_i ~~=~~ \frac{a_p+a_q}{2} \cdot (q-p+1)\]

    Determine the common difference.

    \[d ~=~ (20)-(17) ~=~ 3\]

    Find the value of \(a_{15}\) using the explicit arithmetic formula.

    \[a_{n} = a_1+(n-1)d \] \[a_{15} = 17+(15-1)(3)\] \[a_{15} = 59\]

    So,

    \[\sum_{i=1}^{15} a_i ~=~ \frac{(17)+(59)}{2} \cdot (15)\]

    Evaluate. \[\sum_{i=1}^{15} a_i ~=~ 570\]

    I think of an arithmetic series as representing stacks of coins in a stair pattern. Our goal is to count the coins.

    
    

    Because of the regularity of the step size, the average stack height is simply the average of the first and last stacks.

    If we redistribute the wealth, then all the stacks have the average height.

    
    

    Thus, we just need to multiply the number of stacks by the average height. And the average height is simply the average of the first and last stack. \(38\cdot15 = 570\).

    
    
40. 
    

    Question

    prevnext

    Let \(a_{i}\) represent the \(i\)th term of an arithmetic series, and \[a_1=4\] \[a_2=10\]

    Find the sum of the first 15 terms. In other words, \[\sum_{i=1}^{15}a_i = \,?\]

    
    

    Solution

    Show solution

    The sum of an arithmetic series is found by averaging the first and last terms and multiplying by the number of terms.

    \[\sum_{i=p}^q a_i ~~=~~ \frac{a_p+a_q}{2} \cdot (q-p+1)\]

    Determine the common difference.

    \[d ~=~ (10)-(4) ~=~ 6\]

    Find the value of \(a_{15}\) using the explicit arithmetic formula.

    \[a_{n} = a_1+(n-1)d \] \[a_{15} = 4+(15-1)(6)\] \[a_{15} = 88\]

    So,

    \[\sum_{i=1}^{15} a_i ~=~ \frac{(4)+(88)}{2} \cdot (15)\]

    Evaluate. \[\sum_{i=1}^{15} a_i ~=~ 690\]

    I think of an arithmetic series as representing stacks of coins in a stair pattern. Our goal is to count the coins.

    
    

    Because of the regularity of the step size, the average stack height is simply the average of the first and last stacks.

    If we redistribute the wealth, then all the stacks have the average height.

    
    

    Thus, we just need to multiply the number of stacks by the average height. And the average height is simply the average of the first and last stack. \(46\cdot15 = 690\).

    
    
41. 
    

    Question

    prevnext

    x	y
    49.4	486.7
    53.7	539.6
    52.0	551.3
    57.6	602.5
    57.0	595.0
    48.8	519.4
    51.5	519.0
    58.8	585.1

    Find the trendline using a simple linear regression. Do not assume the \(y\)-intercept is 0.

    Using the linear model, predict \(y\) when \(x=51\).

    
    

    Solution

    Show solution
    • Copy the \(x\) and \(y\) values into a spreadsheet.
    • Highlight the data. Insert chart \(\to\) scatterplot.
    • Under Setup, click Use column A as labels
    • Under Customize \(\to\) Series, click Trendline
    • Change the Label to Use Equation

    The chart+trendline provide a nice visual, and an approximate linear model. To get more exact values for the slope and y-intercept, use the =linest() function. The first argument is the y values, and the second argument is the x values. For the third argument (calculate_b), use true. For the fourth argument (verbose), use false.

    	A	B	C
    1	x	y
    2	49.4	486.7	=linest(B2:B9, A2:A9, TRUE, FALSE)
    3	53.7	539.6
    4	52	551.3
    5	57.6	602.5
    6	57	595
    7	48.8	519.4
    8	51.5	519
    9	58.8	585.1

    
    

    That linest will provide two numbers: the slope 9.9769216 and the intercept 15.0620037.

    So, our linear model is: \[y = (9.9769216)x + (15.0620037) \]

    Plug in the given value of \(x=51\) for prediction.

    \[y ~=~ (9.9769216)(51) + (15.0620037) ~=~ 523.8850039\]

    

    plot of chunk unnamed-chunk-1

    
    
42. 
    

    Question

    prevnext

    x	y
    85.3	646.5
    89.7	732.7
    78.5	597.1
    91.4	702.4
    90.3	715.2
    76.2	589.7
    80.7	649.5

    Find the trendline using a simple linear regression. Do not assume the \(y\)-intercept is 0.

    Using the linear model, predict \(y\) when \(x=71\).

    
    

    Solution

    Show solution
    • Copy the \(x\) and \(y\) values into a spreadsheet.
    • Highlight the data. Insert chart \(\to\) scatterplot.
    • Under Setup, click Use column A as labels
    • Under Customize \(\to\) Series, click Trendline
    • Change the Label to Use Equation

    The chart+trendline provide a nice visual, and an approximate linear model. To get more exact values for the slope and y-intercept, use the =linest() function. The first argument is the y values, and the second argument is the x values. For the third argument (calculate_b), use true. For the fourth argument (verbose), use false.

    	A	B	C
    1	x	y
    2	85.3	646.5	=linest(B2:B8, A2:A8, TRUE, FALSE)
    3	89.7	732.7
    4	78.5	597.1
    5	91.4	702.4
    6	90.3	715.2
    7	76.2	589.7
    8	80.7	649.5

    
    

    That linest will provide two numbers: the slope 8.6746397 and the intercept -71.8791701.

    So, our linear model is: \[y = (8.6746397)x + (-71.8791701) \]

    Plug in the given value of \(x=71\) for prediction.

    \[y ~=~ (8.6746397)(71) + (-71.8791701) ~=~ 544.0202515\]

    

    plot of chunk unnamed-chunk-1

    
    
43. 
    

    Question

    prevnext

    x	y
    26.4	92.5
    27.7	97.7
    22.3	79.6
    24.4	89.1
    26.2	91.9
    25.4	90.8
    29.3	98.9

    Find the trendline using a simple linear regression. Do not assume the \(y\)-intercept is 0.

    Using the linear model, predict \(y\) when \(x=13\).

    
    

    Solution

    Show solution
    • Copy the \(x\) and \(y\) values into a spreadsheet.
    • Highlight the data. Insert chart \(\to\) scatterplot.
    • Under Setup, click Use column A as labels
    • Under Customize \(\to\) Series, click Trendline
    • Change the Label to Use Equation

    The chart+trendline provide a nice visual, and an approximate linear model. To get more exact values for the slope and y-intercept, use the =linest() function. The first argument is the y values, and the second argument is the x values. For the third argument (calculate_b), use true. For the fourth argument (verbose), use false.

    	A	B	C
    1	x	y
    2	26.4	92.5	=linest(B2:B8, A2:A8, TRUE, FALSE)
    3	27.7	97.7
    4	22.3	79.6
    5	24.4	89.1
    6	26.2	91.9
    7	25.4	90.8
    8	29.3	98.9

    
    

    That linest will provide two numbers: the slope 2.7383199 and the intercept 20.4210381.

    So, our linear model is: \[y = (2.7383199)x + (20.4210381) \]

    Plug in the given value of \(x=13\) for prediction.

    \[y ~=~ (2.7383199)(13) + (20.4210381) ~=~ 56.0191973\]

    

    plot of chunk unnamed-chunk-1

    
    
44. 
    

    Question

    prevnext

    x	y
    25.6	140.8
    24.1	127.7
    22.4	115.8
    23.8	134.7
    22.0	122.0
    26.3	144.8
    24.2	124.0
    21.7	122.6

    Find the trendline using a simple linear regression. Do not assume the \(y\)-intercept is 0.

    Using the linear model, predict \(y\) when \(x=35\).

    
    

    Solution

    Show solution
    • Copy the \(x\) and \(y\) values into a spreadsheet.
    • Highlight the data. Insert chart \(\to\) scatterplot.
    • Under Setup, click Use column A as labels
    • Under Customize \(\to\) Series, click Trendline
    • Change the Label to Use Equation

    The chart+trendline provide a nice visual, and an approximate linear model. To get more exact values for the slope and y-intercept, use the =linest() function. The first argument is the y values, and the second argument is the x values. For the third argument (calculate_b), use true. For the fourth argument (verbose), use false.

    	A	B	C
    1	x	y
    2	25.6	140.8	=linest(B2:B9, A2:A9, TRUE, FALSE)
    3	24.1	127.7
    4	22.4	115.8
    5	23.8	134.7
    6	22	122
    7	26.3	144.8
    8	24.2	124
    9	21.7	122.6

    
    

    That linest will provide two numbers: the slope 5.3201474 and the intercept 2.6299981.

    So, our linear model is: \[y = (5.3201474)x + (2.6299981) \]

    Plug in the given value of \(x=35\) for prediction.

    \[y ~=~ (5.3201474)(35) + (2.6299981) ~=~ 188.8351561\]

    

    plot of chunk unnamed-chunk-1

    
    
45. 
    

    Question

    prevnext

    x	y
    35.1	373.7
    34.9	365.5
    33.6	343.2
    42.0	429.4
    41.5	475.1
    31.5	333.3
    30.0	350.1
    29.1	316.2

    Find the trendline using a simple linear regression. Do not assume the \(y\)-intercept is 0.

    Using the linear model, predict \(y\) when \(x=36\).

    
    

    Solution

    Show solution
    • Copy the \(x\) and \(y\) values into a spreadsheet.
    • Highlight the data. Insert chart \(\to\) scatterplot.
    • Under Setup, click Use column A as labels
    • Under Customize \(\to\) Series, click Trendline
    • Change the Label to Use Equation

    The chart+trendline provide a nice visual, and an approximate linear model. To get more exact values for the slope and y-intercept, use the =linest() function. The first argument is the y values, and the second argument is the x values. For the third argument (calculate_b), use true. For the fourth argument (verbose), use false.

    	A	B	C
    1	x	y
    2	35.1	373.7	=linest(B2:B9, A2:A9, TRUE, FALSE)
    3	34.9	365.5
    4	33.6	343.2
    5	42	429.4
    6	41.5	475.1
    7	31.5	333.3
    8	30	350.1
    9	29.1	316.2

    
    

    That linest will provide two numbers: the slope 10.2672149 and the intercept 16.9118038.

    So, our linear model is: \[y = (10.2672149)x + (16.9118038) \]

    Plug in the given value of \(x=36\) for prediction.

    \[y ~=~ (10.2672149)(36) + (16.9118038) ~=~ 386.5315391\]

    

    plot of chunk unnamed-chunk-1

    
    
46. 
    

    Question

    prevnext

    x	y
    36.3	210.2
    29.2	199.6
    36.8	240.0
    27.9	189.6
    35.7	236.2
    35.1	212.2
    35.0	213.0
    28.9	188.4

    Find the trendline using a simple linear regression. Do not assume the \(y\)-intercept is 0.

    Using the linear model, predict \(y\) when \(x=42\).

    
    

    Solution

    Show solution
    • Copy the \(x\) and \(y\) values into a spreadsheet.
    • Highlight the data. Insert chart \(\to\) scatterplot.
    • Under Setup, click Use column A as labels
    • Under Customize \(\to\) Series, click Trendline
    • Change the Label to Use Equation

    The chart+trendline provide a nice visual, and an approximate linear model. To get more exact values for the slope and y-intercept, use the =linest() function. The first argument is the y values, and the second argument is the x values. For the third argument (calculate_b), use true. For the fourth argument (verbose), use false.

    	A	B	C
    1	x	y
    2	36.3	210.2	=linest(B2:B9, A2:A9, TRUE, FALSE)
    3	29.2	199.6
    4	36.8	240
    5	27.9	189.6
    6	35.7	236.2
    7	35.1	212.2
    8	35	213
    9	28.9	188.4

    
    

    That linest will provide two numbers: the slope 4.3498364 and the intercept 67.1160418.

    So, our linear model is: \[y = (4.3498364)x + (67.1160418) \]

    Plug in the given value of \(x=42\) for prediction.

    \[y ~=~ (4.3498364)(42) + (67.1160418) ~=~ 249.8091711\]

    

    plot of chunk unnamed-chunk-1

    
    
47. 
    

    Question

    prevnext

    x	y
    60.0	367.4
    59.0	373.8
    57.9	356.4
    62.9	381.2
    56.9	364.5
    58.7	359.5

    Find the trendline using a simple linear regression. Do not assume the \(y\)-intercept is 0.

    Using the linear model, predict \(y\) when \(x=47\).

    
    

    Solution

    Show solution
    • Copy the \(x\) and \(y\) values into a spreadsheet.
    • Highlight the data. Insert chart \(\to\) scatterplot.
    • Under Setup, click Use column A as labels
    • Under Customize \(\to\) Series, click Trendline
    • Change the Label to Use Equation

    The chart+trendline provide a nice visual, and an approximate linear model. To get more exact values for the slope and y-intercept, use the =linest() function. The first argument is the y values, and the second argument is the x values. For the third argument (calculate_b), use true. For the fourth argument (verbose), use false.

    	A	B	C
    1	x	y
    2	60	367.4	=linest(B2:B7, A2:A7, TRUE, FALSE)
    3	59	373.8
    4	57.9	356.4
    5	62.9	381.2
    6	56.9	364.5
    7	58.7	359.5

    
    

    That linest will provide two numbers: the slope 3.4618709 and the intercept 162.0751775.

    So, our linear model is: \[y = (3.4618709)x + (162.0751775) \]

    Plug in the given value of \(x=47\) for prediction.

    \[y ~=~ (3.4618709)(47) + (162.0751775) ~=~ 324.7831121\]

    

    plot of chunk unnamed-chunk-1

    
    
48. 
    

    Question

    prevnext

    x	y
    16.2	111.9
    21.1	134.7
    13.2	79.0
    14.4	83.4
    13.2	86.2
    21.2	126.7
    19.3	124.6
    20.6	134.1
    23.7	131.8

    Find the trendline using a simple linear regression. Do not assume the \(y\)-intercept is 0.

    Using the linear model, predict \(y\) when \(x=3\).

    
    

    Solution

    Show solution
    • Copy the \(x\) and \(y\) values into a spreadsheet.
    • Highlight the data. Insert chart \(\to\) scatterplot.
    • Under Setup, click Use column A as labels
    • Under Customize \(\to\) Series, click Trendline
    • Change the Label to Use Equation

    The chart+trendline provide a nice visual, and an approximate linear model. To get more exact values for the slope and y-intercept, use the =linest() function. The first argument is the y values, and the second argument is the x values. For the third argument (calculate_b), use true. For the fourth argument (verbose), use false.

    	A	B	C
    1	x	y
    2	16.2	111.9	=linest(B2:B10, A2:A10, TRUE, FALSE)
    3	21.1	134.7
    4	13.2	79
    5	14.4	83.4
    6	13.2	86.2
    7	21.2	126.7
    8	19.3	124.6
    9	20.6	134.1
    10	23.7	131.8

    
    

    That linest will provide two numbers: the slope 5.6029436 and the intercept 11.0756103.

    So, our linear model is: \[y = (5.6029436)x + (11.0756103) \]

    Plug in the given value of \(x=3\) for prediction.

    \[y ~=~ (5.6029436)(3) + (11.0756103) ~=~ 27.884441\]

    

    plot of chunk unnamed-chunk-1

    
    
49. 
    

    Question

    prevnext

    x	y
    40.7	297.6
    42.4	311.8
    42.3	306.7
    39.7	296.9
    41.2	307.7
    43.2	310.6
    44.2	326.6
    44.2	319.5

    Find the trendline using a simple linear regression. Do not assume the \(y\)-intercept is 0.

    Using the linear model, predict \(y\) when \(x=36\).

    
    

    Solution

    Show solution
    • Copy the \(x\) and \(y\) values into a spreadsheet.
    • Highlight the data. Insert chart \(\to\) scatterplot.
    • Under Setup, click Use column A as labels
    • Under Customize \(\to\) Series, click Trendline
    • Change the Label to Use Equation

    The chart+trendline provide a nice visual, and an approximate linear model. To get more exact values for the slope and y-intercept, use the =linest() function. The first argument is the y values, and the second argument is the x values. For the third argument (calculate_b), use true. For the fourth argument (verbose), use false.

    	A	B	C
    1	x	y
    2	40.7	297.6	=linest(B2:B9, A2:A9, TRUE, FALSE)
    3	42.4	311.8
    4	42.3	306.7
    5	39.7	296.9
    6	41.2	307.7
    7	43.2	310.6
    8	44.2	326.6
    9	44.2	319.5

    
    

    That linest will provide two numbers: the slope 5.7489043 and the intercept 66.8556537.

    So, our linear model is: \[y = (5.7489043)x + (66.8556537) \]

    Plug in the given value of \(x=36\) for prediction.

    \[y ~=~ (5.7489043)(36) + (66.8556537) ~=~ 273.8162093\]

    

    plot of chunk unnamed-chunk-1

    
    
50. 
    

    Question

    prevnext

    x	y
    24.4	162.0
    24.2	158.8
    22.5	142.9
    21.7	144.2
    23.8	157.5
    22.5	141.2
    27.2	175.7
    26.9	172.0
    24.3	158.9

    Find the trendline using a simple linear regression. Do not assume the \(y\)-intercept is 0.

    Using the linear model, predict \(y\) when \(x=23\).

    
    

    Solution

    Show solution
    • Copy the \(x\) and \(y\) values into a spreadsheet.
    • Highlight the data. Insert chart \(\to\) scatterplot.
    • Under Setup, click Use column A as labels
    • Under Customize \(\to\) Series, click Trendline
    • Change the Label to Use Equation

    The chart+trendline provide a nice visual, and an approximate linear model. To get more exact values for the slope and y-intercept, use the =linest() function. The first argument is the y values, and the second argument is the x values. For the third argument (calculate_b), use true. For the fourth argument (verbose), use false.

    	A	B	C
    1	x	y
    2	24.4	162	=linest(B2:B10, A2:A10, TRUE, FALSE)
    3	24.2	158.8
    4	22.5	142.9
    5	21.7	144.2
    6	23.8	157.5
    7	22.5	141.2
    8	27.2	175.7
    9	26.9	172
    10	24.3	158.9

    
    

    That linest will provide two numbers: the slope 6.3263207 and the intercept 4.1361384.

    So, our linear model is: \[y = (6.3263207)x + (4.1361384) \]

    Plug in the given value of \(x=23\) for prediction.

    \[y ~=~ (6.3263207)(23) + (4.1361384) ~=~ 149.6415147\]

    

    plot of chunk unnamed-chunk-1

    
    
51. 
    

    Question

    prevnext

    Solve the system: \[y = 8 x + 4\] \[y = 3 x - 1\]

    Your answers:

    • \(x=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    • \(y=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    
    
    
    

    Solution

    Show solution

    Eliminate \(y\) by subtrating the second equation from the first. \[0 = 5 x + 5\]

    Subtract 5 from both sides.

    \[-5 = 5 x\]

    Multiply/divide to solve for \(x\).

    \[x = -1\]

    Plug in -1 for \(x\) in either original equation.

    \[y = 8 (-1) + 4\]

    \[y = -4\]

    You can also solve this by graphing!

    
    

    
    
52. 
    

    Question

    prevnext

    Solve the system: \[y = - 3 x - 3\] \[y = - 2 x - 1\]

    Your answers:

    • \(x=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    • \(y=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    
    
    
    

    Solution

    Show solution

    Eliminate \(y\) by subtrating the second equation from the first. \[0 = - x - 2\]

    Subtract -2 from both sides.

    \[2 = - x\]

    Multiply/divide to solve for \(x\).

    \[x = -2\]

    Plug in -2 for \(x\) in either original equation.

    \[y = - 3 (-2) - 3\]

    \[y = 3\]

    You can also solve this by graphing!

    
    

    
    
53. 
    

    Question

    prevnext

    Solve the system: \[y = - 2 x - 7\] \[y = - \frac{x}{2} - 4\]

    Your answers:

    • \(x=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    • \(y=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    
    
    
    

    Solution

    Show solution

    Eliminate \(y\) by subtrating the second equation from the first. \[0 = - \frac{3 x}{2} - 3\]

    Subtract -3 from both sides.

    \[3 = - \frac{3 x}{2}\]

    Multiply/divide to solve for \(x\).

    \[x = -2\]

    Plug in -2 for \(x\) in either original equation.

    \[y = - 2 (-2) - 7\]

    \[y = -3\]

    You can also solve this by graphing!

    
    

    
    
54. 
    

    Question

    prevnext

    Solve the system: \[y = 7 - 7 x\] \[y = - \frac{5 x}{2} - 2\]

    Your answers:

    • \(x=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    • \(y=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    
    
    
    

    Solution

    Show solution

    Eliminate \(y\) by subtrating the second equation from the first. \[0 = 9 - \frac{9 x}{2}\]

    Subtract 9 from both sides.

    \[-9 = - \frac{9 x}{2}\]

    Multiply/divide to solve for \(x\).

    \[x = 2\]

    Plug in 2 for \(x\) in either original equation.

    \[y = 7 - 7 (2)\]

    \[y = -7\]

    You can also solve this by graphing!

    
    

    
    
55. 
    

    Question

    prevnext

    Solve the system: \[y = \frac{4 x}{5} - 7\] \[y = 7 - 2 x\]

    Your answers:

    • \(x=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    • \(y=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    
    
    
    

    Solution

    Show solution

    Eliminate \(y\) by subtrating the second equation from the first. \[0 = \frac{14 x}{5} - 14\]

    Subtract -14 from both sides.

    \[14 = \frac{14 x}{5}\]

    Multiply/divide to solve for \(x\).

    \[x = 5\]

    Plug in 5 for \(x\) in either original equation.

    \[y = \frac{4 (5)}{5} - 7\]

    \[y = -3\]

    You can also solve this by graphing!

    
    

    
    
56. 
    

    Question

    prevnext

    Solve the system: \[y = \frac{x}{3} + 4\] \[y = 8 - x\]

    Your answers:

    • \(x=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    • \(y=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    
    
    
    

    Solution

    Show solution

    Eliminate \(y\) by subtrating the second equation from the first. \[0 = \frac{4 x}{3} - 4\]

    Subtract -4 from both sides.

    \[4 = \frac{4 x}{3}\]

    Multiply/divide to solve for \(x\).

    \[x = 3\]

    Plug in 3 for \(x\) in either original equation.

    \[y = \frac{(3)}{3} + 4\]

    \[y = 5\]

    You can also solve this by graphing!

    
    

    
    
57. 
    

    Question

    prevnext

    Solve the system: \[y = 2 x + 2\] \[y = x + 4\]

    Your answers:

    • \(x=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    • \(y=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    
    
    
    

    Solution

    Show solution

    Eliminate \(y\) by subtrating the second equation from the first. \[0 = x - 2\]

    Subtract -2 from both sides.

    \[2 = x\]

    Multiply/divide to solve for \(x\).

    \[x = 2\]

    Plug in 2 for \(x\) in either original equation.

    \[y = 2 (2) + 2\]

    \[y = 6\]

    You can also solve this by graphing!

    
    

    
    
58. 
    

    Question

    prevnext

    Solve the system: \[y = \frac{7 x}{4} - 6\] \[y = 3 - \frac{x}{2}\]

    Your answers:

    • \(x=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    • \(y=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    
    
    
    

    Solution

    Show solution

    Eliminate \(y\) by subtrating the second equation from the first. \[0 = \frac{9 x}{4} - 9\]

    Subtract -9 from both sides.

    \[9 = \frac{9 x}{4}\]

    Multiply/divide to solve for \(x\).

    \[x = 4\]

    Plug in 4 for \(x\) in either original equation.

    \[y = \frac{7 (4)}{4} - 6\]

    \[y = 1\]

    You can also solve this by graphing!

    
    

    
    
59. 
    

    Question

    prevnext

    Solve the system: \[y = \frac{x}{2} - 8\] \[y = 7 x + 5\]

    Your answers:

    • \(x=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    • \(y=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    
    
    
    

    Solution

    Show solution

    Eliminate \(y\) by subtrating the second equation from the first. \[0 = - \frac{13 x}{2} - 13\]

    Subtract -13 from both sides.

    \[13 = - \frac{13 x}{2}\]

    Multiply/divide to solve for \(x\).

    \[x = -2\]

    Plug in -2 for \(x\) in either original equation.

    \[y = \frac{(-2)}{2} - 8\]

    \[y = -9\]

    You can also solve this by graphing!

    
    

    
    
60. 
    

    Question

    prevnext

    Solve the system: \[y = x - 2\] \[y = - x - 6\]

    Your answers:

    • \(x=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    • \(y=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    
    
    
    

    Solution

    Show solution

    Eliminate \(y\) by subtrating the second equation from the first. \[0 = 2 x + 4\]

    Subtract 4 from both sides.

    \[-4 = 2 x\]

    Multiply/divide to solve for \(x\).

    \[x = -2\]

    Plug in -2 for \(x\) in either original equation.

    \[y = (-2) - 2\]

    \[y = -4\]

    You can also solve this by graphing!

    
    

    
    
61. 
    

    Question

    prevnext

    Solve the system: \[4 x - 3 y = 12\] \[x - y = 2\]

    Your answers:

    • \(x=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    • \(y=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    
    
    
    

    Solution

    Show solution

    When graphing linear equations given in standard form, it is typically best to determine both the \(x\)-intercept and the \(y\)-intercept.

    To find the \(x\)-intercepts, set \(y\) to 0. \[4x_{\text{int}1}+(-3)(0) = 12 \] \[4x_{\text{int}1} = 12 \] \[x_{\text{int}1} = \frac{12}{4} = 3 \]

    \[1x_{\text{int}2}+(-1)(0) = 2 \] \[1x_{\text{int}2} = 2 \] \[x_{\text{int}2} = \frac{2}{1} = 2 \]

    To find the \(y\)-intercepts, set \(x\) to 0. \[(4)(0)+(-3)(y_{\text{int1}}) = 12 \] \[-3y_{\text{int}1} = 12 \] \[y_{\text{int}1} = \frac{12}{-3} = -4 \]

    \[(1)(0)+(-1)y_{\text{int2}} = 2 \] \[-1y_{\text{int}2} = 2 \] \[y_{\text{int}2} = \frac{2}{-1} = -2 \]

    Plot those intercepts, determine slope pattern, draw lines, and find intersection.

    
    

    The intersection occurs at \((6,\,4)\).

    
    
62. 
    

    Question

    prevnext

    Solve the system: \[- 3 x + y = 6\] \[- x + y = 8\]

    Your answers:

    • \(x=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    • \(y=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    
    
    
    

    Solution

    Show solution

    When graphing linear equations given in standard form, it is typically best to determine both the \(x\)-intercept and the \(y\)-intercept.

    To find the \(x\)-intercepts, set \(y\) to 0. \[-3x_{\text{int}1}+(1)(0) = 6 \] \[-3x_{\text{int}1} = 6 \] \[x_{\text{int}1} = \frac{6}{-3} = -2 \]

    \[-1x_{\text{int}2}+(1)(0) = 8 \] \[-1x_{\text{int}2} = 8 \] \[x_{\text{int}2} = \frac{8}{-1} = -8 \]

    To find the \(y\)-intercepts, set \(x\) to 0. \[(-3)(0)+(1)(y_{\text{int1}}) = 6 \] \[1y_{\text{int}1} = 6 \] \[y_{\text{int}1} = \frac{6}{1} = 6 \]

    \[(-1)(0)+(1)y_{\text{int2}} = 8 \] \[1y_{\text{int}2} = 8 \] \[y_{\text{int}2} = \frac{8}{1} = 8 \]

    Plot those intercepts, determine slope pattern, draw lines, and find intersection.

    
    

    The intersection occurs at \((1,\,9)\).

    
    
63. 
    

    Question

    prevnext

    Solve the system: \[x + 3 y = 3\] \[- 2 x + 3 y = 12\]

    Your answers:

    • \(x=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    • \(y=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    
    
    
    

    Solution

    Show solution

    When graphing linear equations given in standard form, it is typically best to determine both the \(x\)-intercept and the \(y\)-intercept.

    To find the \(x\)-intercepts, set \(y\) to 0. \[1x_{\text{int}1}+(3)(0) = 3 \] \[1x_{\text{int}1} = 3 \] \[x_{\text{int}1} = \frac{3}{1} = 3 \]

    \[-2x_{\text{int}2}+(3)(0) = 12 \] \[-2x_{\text{int}2} = 12 \] \[x_{\text{int}2} = \frac{12}{-2} = -6 \]

    To find the \(y\)-intercepts, set \(x\) to 0. \[(1)(0)+(3)(y_{\text{int1}}) = 3 \] \[3y_{\text{int}1} = 3 \] \[y_{\text{int}1} = \frac{3}{3} = 1 \]

    \[(-2)(0)+(3)y_{\text{int2}} = 12 \] \[3y_{\text{int}2} = 12 \] \[y_{\text{int}2} = \frac{12}{3} = 4 \]

    Plot those intercepts, determine slope pattern, draw lines, and find intersection.

    
    

    The intersection occurs at \((-3,\,2)\).

    
    
64. 
    

    Question

    prevnext

    Solve the system: \[x - y = 3\] \[- 7 x + 2 y = 14\]

    Your answers:

    • \(x=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    • \(y=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    
    
    
    

    Solution

    Show solution

    When graphing linear equations given in standard form, it is typically best to determine both the \(x\)-intercept and the \(y\)-intercept.

    To find the \(x\)-intercepts, set \(y\) to 0. \[1x_{\text{int}1}+(-1)(0) = 3 \] \[1x_{\text{int}1} = 3 \] \[x_{\text{int}1} = \frac{3}{1} = 3 \]

    \[-7x_{\text{int}2}+(2)(0) = 14 \] \[-7x_{\text{int}2} = 14 \] \[x_{\text{int}2} = \frac{14}{-7} = -2 \]

    To find the \(y\)-intercepts, set \(x\) to 0. \[(1)(0)+(-1)(y_{\text{int1}}) = 3 \] \[-1y_{\text{int}1} = 3 \] \[y_{\text{int}1} = \frac{3}{-1} = -3 \]

    \[(-7)(0)+(2)y_{\text{int2}} = 14 \] \[2y_{\text{int}2} = 14 \] \[y_{\text{int}2} = \frac{14}{2} = 7 \]

    Plot those intercepts, determine slope pattern, draw lines, and find intersection.

    
    

    The intersection occurs at \((-4,\,-7)\).

    
    
65. 
    

    Question

    prevnext

    Solve the system: \[x - 3 y = 9\] \[2 x - 3 y = 12\]

    Your answers:

    • \(x=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    • \(y=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    
    
    
    

    Solution

    Show solution

    When graphing linear equations given in standard form, it is typically best to determine both the \(x\)-intercept and the \(y\)-intercept.

    To find the \(x\)-intercepts, set \(y\) to 0. \[1x_{\text{int}1}+(-3)(0) = 9 \] \[1x_{\text{int}1} = 9 \] \[x_{\text{int}1} = \frac{9}{1} = 9 \]

    \[2x_{\text{int}2}+(-3)(0) = 12 \] \[2x_{\text{int}2} = 12 \] \[x_{\text{int}2} = \frac{12}{2} = 6 \]

    To find the \(y\)-intercepts, set \(x\) to 0. \[(1)(0)+(-3)(y_{\text{int1}}) = 9 \] \[-3y_{\text{int}1} = 9 \] \[y_{\text{int}1} = \frac{9}{-3} = -3 \]

    \[(2)(0)+(-3)y_{\text{int2}} = 12 \] \[-3y_{\text{int}2} = 12 \] \[y_{\text{int}2} = \frac{12}{-3} = -4 \]

    Plot those intercepts, determine slope pattern, draw lines, and find intersection.

    
    

    The intersection occurs at \((3,\,-2)\).

    
    
66. 
    

    Question

    prevnext

    Solve the system: \[x - 2 y = 4\] \[- 5 x + 3 y = 15\]

    Your answers:

    • \(x=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    • \(y=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    
    
    
    

    Solution

    Show solution

    When graphing linear equations given in standard form, it is typically best to determine both the \(x\)-intercept and the \(y\)-intercept.

    To find the \(x\)-intercepts, set \(y\) to 0. \[1x_{\text{int}1}+(-2)(0) = 4 \] \[1x_{\text{int}1} = 4 \] \[x_{\text{int}1} = \frac{4}{1} = 4 \]

    \[-5x_{\text{int}2}+(3)(0) = 15 \] \[-5x_{\text{int}2} = 15 \] \[x_{\text{int}2} = \frac{15}{-5} = -3 \]

    To find the \(y\)-intercepts, set \(x\) to 0. \[(1)(0)+(-2)(y_{\text{int1}}) = 4 \] \[-2y_{\text{int}1} = 4 \] \[y_{\text{int}1} = \frac{4}{-2} = -2 \]

    \[(-5)(0)+(3)y_{\text{int2}} = 15 \] \[3y_{\text{int}2} = 15 \] \[y_{\text{int}2} = \frac{15}{3} = 5 \]

    Plot those intercepts, determine slope pattern, draw lines, and find intersection.

    
    

    The intersection occurs at \((-6,\,-5)\).

    
    
67. 
    

    Question

    prevnext

    Solve the system: \[- 3 x - 2 y = 6\] \[3 x + y = 3\]

    Your answers:

    • \(x=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    • \(y=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    
    
    
    

    Solution

    Show solution

    When graphing linear equations given in standard form, it is typically best to determine both the \(x\)-intercept and the \(y\)-intercept.

    To find the \(x\)-intercepts, set \(y\) to 0. \[-3x_{\text{int}1}+(-2)(0) = 6 \] \[-3x_{\text{int}1} = 6 \] \[x_{\text{int}1} = \frac{6}{-3} = -2 \]

    \[3x_{\text{int}2}+(1)(0) = 3 \] \[3x_{\text{int}2} = 3 \] \[x_{\text{int}2} = \frac{3}{3} = 1 \]

    To find the \(y\)-intercepts, set \(x\) to 0. \[(-3)(0)+(-2)(y_{\text{int1}}) = 6 \] \[-2y_{\text{int}1} = 6 \] \[y_{\text{int}1} = \frac{6}{-2} = -3 \]

    \[(3)(0)+(1)y_{\text{int2}} = 3 \] \[1y_{\text{int}2} = 3 \] \[y_{\text{int}2} = \frac{3}{1} = 3 \]

    Plot those intercepts, determine slope pattern, draw lines, and find intersection.

    
    

    The intersection occurs at \((4,\,-9)\).

    
    
68. 
    

    Question

    prevnext

    Solve the system: \[x - y = 1\] \[2 x - 3 y = 6\]

    Your answers:

    • \(x=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    • \(y=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    
    
    
    

    Solution

    Show solution

    When graphing linear equations given in standard form, it is typically best to determine both the \(x\)-intercept and the \(y\)-intercept.

    To find the \(x\)-intercepts, set \(y\) to 0. \[1x_{\text{int}1}+(-1)(0) = 1 \] \[1x_{\text{int}1} = 1 \] \[x_{\text{int}1} = \frac{1}{1} = 1 \]

    \[2x_{\text{int}2}+(-3)(0) = 6 \] \[2x_{\text{int}2} = 6 \] \[x_{\text{int}2} = \frac{6}{2} = 3 \]

    To find the \(y\)-intercepts, set \(x\) to 0. \[(1)(0)+(-1)(y_{\text{int1}}) = 1 \] \[-1y_{\text{int}1} = 1 \] \[y_{\text{int}1} = \frac{1}{-1} = -1 \]

    \[(2)(0)+(-3)y_{\text{int2}} = 6 \] \[-3y_{\text{int}2} = 6 \] \[y_{\text{int}2} = \frac{6}{-3} = -2 \]

    Plot those intercepts, determine slope pattern, draw lines, and find intersection.

    
    

    The intersection occurs at \((-3,\,-4)\).

    
    
69. 
    

    Question

    prevnext

    Solve the system: \[- x - 2 y = 2\] \[3 x - 2 y = 18\]

    Your answers:

    • \(x=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    • \(y=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    
    
    
    

    Solution

    Show solution

    When graphing linear equations given in standard form, it is typically best to determine both the \(x\)-intercept and the \(y\)-intercept.

    To find the \(x\)-intercepts, set \(y\) to 0. \[-1x_{\text{int}1}+(-2)(0) = 2 \] \[-1x_{\text{int}1} = 2 \] \[x_{\text{int}1} = \frac{2}{-1} = -2 \]

    \[3x_{\text{int}2}+(-2)(0) = 18 \] \[3x_{\text{int}2} = 18 \] \[x_{\text{int}2} = \frac{18}{3} = 6 \]

    To find the \(y\)-intercepts, set \(x\) to 0. \[(-1)(0)+(-2)(y_{\text{int1}}) = 2 \] \[-2y_{\text{int}1} = 2 \] \[y_{\text{int}1} = \frac{2}{-2} = -1 \]

    \[(3)(0)+(-2)y_{\text{int2}} = 18 \] \[-2y_{\text{int}2} = 18 \] \[y_{\text{int}2} = \frac{18}{-2} = -9 \]

    Plot those intercepts, determine slope pattern, draw lines, and find intersection.

    
    

    The intersection occurs at \((4,\,-3)\).

    
    
70. 
    

    Question

    prevnext

    Solve the system: \[- x + y = 3\] \[- 4 x + 5 y = 20\]

    Your answers:

    • \(x=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    • \(y=\) -------- -10-9-8-7-6-5-4-3-2-1012345678910
    
    
    
    

    Solution

    Show solution

    When graphing linear equations given in standard form, it is typically best to determine both the \(x\)-intercept and the \(y\)-intercept.

    To find the \(x\)-intercepts, set \(y\) to 0. \[-1x_{\text{int}1}+(1)(0) = 3 \] \[-1x_{\text{int}1} = 3 \] \[x_{\text{int}1} = \frac{3}{-1} = -3 \]

    \[-4x_{\text{int}2}+(5)(0) = 20 \] \[-4x_{\text{int}2} = 20 \] \[x_{\text{int}2} = \frac{20}{-4} = -5 \]

    To find the \(y\)-intercepts, set \(x\) to 0. \[(-1)(0)+(1)(y_{\text{int1}}) = 3 \] \[1y_{\text{int}1} = 3 \] \[y_{\text{int}1} = \frac{3}{1} = 3 \]

    \[(-4)(0)+(5)y_{\text{int2}} = 20 \] \[5y_{\text{int}2} = 20 \] \[y_{\text{int}2} = \frac{20}{5} = 4 \]

    Plot those intercepts, determine slope pattern, draw lines, and find intersection.

    
    

    The intersection occurs at \((5,\,8)\).

    
    
71. 
    

    Question

    prevnext

    A thief is stealing xots and yivs. Each xot has a mass of 5 kilograms and a volume of 10 liters. Each yiv has a mass of 20 kilograms and a volume of 15 liters. The thief can carry a maximum mass of 100 kilograms and a maximum volume of 150 liters. The profit from each xot is $2.31 and the profit from each yiv is $4.8.

    For your convenience, those numbers are organized in the table below.

    attribute	xot	yiv	capacity
    mass (kg)	5	20	100
    volume (L)	10	15	150
    profit ($)	2.31	4.80	\(\infty\)

    What is the maximum profit the thief can produce? (In dollars.)

    
    

    Solution

    Show solution

    This is an example of linear programming.

    Define \(x\) as number of xots and \(y\) as number of yivs. Write the constraints as inequalities.

    • Mass constraint: \(5x+20y \le 100\)
    • Volume constraint: \(10x+15y \le 150\)

    We can assume that optimal profit will occur at a boundary, so we can just consider the equations. \[5x+20y = 100\] \[10x+15y = 150\]

    If either equation has coefficients that all share a common factor, we can simplify by dividing both sides by the greatest common factor:

    \[1x+4y = 20\] \[2x+3y = 30\]

    We need to find the \(x\) intercepts and the \(y\) intercepts.

    To find the \(x\)-intercepts, set \(y\) to 0. \[1x_{\text{int}1}+(4)(0) = 20 \] \[1x_{\text{int}1} = 20 \] \[x_{\text{int}1} = \frac{20}{1} = 20 \]

    \[2x_{\text{int}2}+(3)(0) = 30 \] \[2x_{\text{int}2} = 30 \] \[x_{\text{int}2} = \frac{30}{2} = 15 \]

    To find the \(y\)-intercepts, set \(x\) to 0. \[(1)(0)+(4)(y_{\text{int1}}) = 20 \] \[4y_{\text{int}1} = 20 \] \[y_{\text{int}1} = \frac{20}{4} = 5 \]

    \[(2)(0)+(3)y_{\text{int2}} = 30 \] \[3y_{\text{int}2} = 30 \] \[y_{\text{int}2} = \frac{30}{3} = 10 \]

    To find the intersection, I would recommend elimination. Multiply the equations to have matching coefficients for \(y\) terms.

    \[3x + 12y = 60\] \[8x + 12y = 120\] Subtract equations.

    \[5x + (0)y = 60\]

    Simplify.

    \[5x = 60\]

    Divide by 5.

    \[x = 12\]

    Plug \(x = 12\) into the first equation. Solve for \(y\).

    \[(1)(12)+4y = 20\] \[12+4y = 20\] \[4y = 8\] \[y = 2\]

    So, the intersection occurs at \((12,\,2)\).

    You can plot the boundaries and shade the feasible region.

    
    

    The maximum profit could occur at \((0,5)\) or \((12,2)\) or \((15,0)\).

    Find those profits, using \(P(x,y)=2.31x+4.8y\).

    \[P(0,5)=(2.31)(0)+(4.8)(5) = 24\] \[P(12,2)=(2.31)(12)+(4.8)(2) = 37.32\] \[P(15,0)=(2.31)(15)+(4.8)(0) = 34.65\]

    So, the maximum profit is $37.32.

    It might help to visualize this in 3D, with xots, yivs, and profit on the 3 axes. Sorry, I was not able to make a 3D plot show up here.

    
    
72. 
    

    Question

    prevnext

    A thief is stealing xots and yivs. Each xot has a mass of 4 kilograms and a volume of 7 liters. Each yiv has a mass of 16 kilograms and a volume of 14 liters. The thief can carry a maximum mass of 64 kilograms and a maximum volume of 98 liters. The profit from each xot is $1.98 and the profit from each yiv is $5.64.

    For your convenience, those numbers are organized in the table below.

    attribute	xot	yiv	capacity
    mass (kg)	4	16	64
    volume (L)	7	14	98
    profit ($)	1.98	5.64	\(\infty\)

    What is the maximum profit the thief can produce? (In dollars.)

    
    

    Solution

    Show solution

    This is an example of linear programming.

    Define \(x\) as number of xots and \(y\) as number of yivs. Write the constraints as inequalities.

    • Mass constraint: \(4x+16y \le 64\)
    • Volume constraint: \(7x+14y \le 98\)

    We can assume that optimal profit will occur at a boundary, so we can just consider the equations. \[4x+16y = 64\] \[7x+14y = 98\]

    If either equation has coefficients that all share a common factor, we can simplify by dividing both sides by the greatest common factor:

    \[1x+4y = 16\] \[1x+2y = 14\]

    We need to find the \(x\) intercepts and the \(y\) intercepts.

    To find the \(x\)-intercepts, set \(y\) to 0. \[1x_{\text{int}1}+(4)(0) = 16 \] \[1x_{\text{int}1} = 16 \] \[x_{\text{int}1} = \frac{16}{1} = 16 \]

    \[1x_{\text{int}2}+(2)(0) = 14 \] \[1x_{\text{int}2} = 14 \] \[x_{\text{int}2} = \frac{14}{1} = 14 \]

    To find the \(y\)-intercepts, set \(x\) to 0. \[(1)(0)+(4)(y_{\text{int1}}) = 16 \] \[4y_{\text{int}1} = 16 \] \[y_{\text{int}1} = \frac{16}{4} = 4 \]

    \[(1)(0)+(2)y_{\text{int2}} = 14 \] \[2y_{\text{int}2} = 14 \] \[y_{\text{int}2} = \frac{14}{2} = 7 \]

    To find the intersection, I would recommend elimination. Multiply the equations to have matching coefficients for \(y\) terms.

    \[1x + 4y = 16\] \[2x + 4y = 28\] Subtract equations.

    \[1x + (0)y = 12\]

    Simplify.

    \[1x = 12\]

    Divide by 1.

    \[x = 12\]

    Plug \(x = 12\) into the first equation. Solve for \(y\).

    \[(1)(12)+4y = 16\] \[12+4y = 16\] \[4y = 4\] \[y = 1\]

    So, the intersection occurs at \((12,\,1)\).

    You can plot the boundaries and shade the feasible region.

    
    

    The maximum profit could occur at \((0,4)\) or \((12,1)\) or \((14,0)\).

    Find those profits, using \(P(x,y)=1.98x+5.64y\).

    \[P(0,4)=(1.98)(0)+(5.64)(4) = 22.56\] \[P(12,1)=(1.98)(12)+(5.64)(1) = 29.4\] \[P(14,0)=(1.98)(14)+(5.64)(0) = 27.72\]

    So, the maximum profit is $29.40.

    It might help to visualize this in 3D, with xots, yivs, and profit on the 3 axes. Sorry, I was not able to make a 3D plot show up here.

    
    
73. 
    

    Question

    prevnext

    A thief is stealing xots and yivs. Each xot has a mass of 6 kilograms and a volume of 5 liters. Each yiv has a mass of 18 kilograms and a volume of 20 liters. The thief can carry a maximum mass of 108 kilograms and a maximum volume of 100 liters. The profit from each xot is $7.77 and the profit from each yiv is $8.81.

    For your convenience, those numbers are organized in the table below.

    attribute	xot	yiv	capacity
    mass (kg)	6	18	108
    volume (L)	5	20	100
    profit ($)	7.77	8.81	\(\infty\)

    What is the maximum profit the thief can produce? (In dollars.)

    
    

    Solution

    Show solution

    This is an example of linear programming.

    Define \(x\) as number of xots and \(y\) as number of yivs. Write the constraints as inequalities.

    • Mass constraint: \(6x+18y \le 108\)
    • Volume constraint: \(5x+20y \le 100\)

    We can assume that optimal profit will occur at a boundary, so we can just consider the equations. \[6x+18y = 108\] \[5x+20y = 100\]

    If either equation has coefficients that all share a common factor, we can simplify by dividing both sides by the greatest common factor:

    \[1x+3y = 18\] \[1x+4y = 20\]

    We need to find the \(x\) intercepts and the \(y\) intercepts.

    To find the \(x\)-intercepts, set \(y\) to 0. \[1x_{\text{int}1}+(3)(0) = 18 \] \[1x_{\text{int}1} = 18 \] \[x_{\text{int}1} = \frac{18}{1} = 18 \]

    \[1x_{\text{int}2}+(4)(0) = 20 \] \[1x_{\text{int}2} = 20 \] \[x_{\text{int}2} = \frac{20}{1} = 20 \]

    To find the \(y\)-intercepts, set \(x\) to 0. \[(1)(0)+(3)(y_{\text{int1}}) = 18 \] \[3y_{\text{int}1} = 18 \] \[y_{\text{int}1} = \frac{18}{3} = 6 \]

    \[(1)(0)+(4)y_{\text{int2}} = 20 \] \[4y_{\text{int}2} = 20 \] \[y_{\text{int}2} = \frac{20}{4} = 5 \]

    To find the intersection, I would recommend elimination. Multiply the equations to have matching coefficients for \(y\) terms.

    \[4x + 12y = 72\] \[3x + 12y = 60\] Subtract equations.

    \[1x + (0)y = 12\]

    Simplify.

    \[1x = 12\]

    Divide by 1.

    \[x = 12\]

    Plug \(x = 12\) into the first equation. Solve for \(y\).

    \[(1)(12)+3y = 18\] \[12+3y = 18\] \[3y = 6\] \[y = 2\]

    So, the intersection occurs at \((12,\,2)\).

    You can plot the boundaries and shade the feasible region.

    
    

    The maximum profit could occur at \((0,5)\) or \((12,2)\) or \((18,0)\).

    Find those profits, using \(P(x,y)=7.77x+8.81y\).

    \[P(0,5)=(7.77)(0)+(8.81)(5) = 44.05\] \[P(12,2)=(7.77)(12)+(8.81)(2) = 110.86\] \[P(18,0)=(7.77)(18)+(8.81)(0) = 139.86\]

    So, the maximum profit is $139.86.

    It might help to visualize this in 3D, with xots, yivs, and profit on the 3 axes. Sorry, I was not able to make a 3D plot show up here.

    
    
74. 
    

    Question

    prevnext

    A thief is stealing xots and yivs. Each xot has a mass of 16 kilograms and a volume of 6 liters. Each yiv has a mass of 8 kilograms and a volume of 18 liters. The thief can carry a maximum mass of 128 kilograms and a maximum volume of 108 liters. The profit from each xot is $7.51 and the profit from each yiv is $8.01.

    For your convenience, those numbers are organized in the table below.

    attribute	xot	yiv	capacity
    mass (kg)	16	8	128
    volume (L)	6	18	108
    profit ($)	7.51	8.01	\(\infty\)

    What is the maximum profit the thief can produce? (In dollars.)

    
    

    Solution

    Show solution

    This is an example of linear programming.

    Define \(x\) as number of xots and \(y\) as number of yivs. Write the constraints as inequalities.

    • Mass constraint: \(16x+8y \le 128\)
    • Volume constraint: \(6x+18y \le 108\)

    We can assume that optimal profit will occur at a boundary, so we can just consider the equations. \[16x+8y = 128\] \[6x+18y = 108\]

    If either equation has coefficients that all share a common factor, we can simplify by dividing both sides by the greatest common factor:

    \[2x+1y = 16\] \[1x+3y = 18\]

    We need to find the \(x\) intercepts and the \(y\) intercepts.

    To find the \(x\)-intercepts, set \(y\) to 0. \[2x_{\text{int}1}+(1)(0) = 16 \] \[2x_{\text{int}1} = 16 \] \[x_{\text{int}1} = \frac{16}{2} = 8 \]

    \[1x_{\text{int}2}+(3)(0) = 18 \] \[1x_{\text{int}2} = 18 \] \[x_{\text{int}2} = \frac{18}{1} = 18 \]

    To find the \(y\)-intercepts, set \(x\) to 0. \[(2)(0)+(1)(y_{\text{int1}}) = 16 \] \[1y_{\text{int}1} = 16 \] \[y_{\text{int}1} = \frac{16}{1} = 16 \]

    \[(1)(0)+(3)y_{\text{int2}} = 18 \] \[3y_{\text{int}2} = 18 \] \[y_{\text{int}2} = \frac{18}{3} = 6 \]

    To find the intersection, I would recommend elimination. Multiply the equations to have matching coefficients for \(y\) terms.

    \[6x + 3y = 48\] \[1x + 3y = 18\] Subtract equations.

    \[5x + (0)y = 30\]

    Simplify.

    \[5x = 30\]

    Divide by 5.

    \[x = 6\]

    Plug \(x = 6\) into the first equation. Solve for \(y\).

    \[(2)(6)+1y = 16\] \[12+1y = 16\] \[1y = 4\] \[y = 4\]

    So, the intersection occurs at \((6,\,4)\).

    You can plot the boundaries and shade the feasible region.

    
    

    The maximum profit could occur at \((0,6)\) or \((6,4)\) or \((8,0)\).

    Find those profits, using \(P(x,y)=7.51x+8.01y\).

    \[P(0,6)=(7.51)(0)+(8.01)(6) = 48.06\] \[P(6,4)=(7.51)(6)+(8.01)(4) = 77.1\] \[P(8,0)=(7.51)(8)+(8.01)(0) = 60.08\]

    So, the maximum profit is $77.10.

    It might help to visualize this in 3D, with xots, yivs, and profit on the 3 axes. Sorry, I was not able to make a 3D plot show up here.

    
    
75. 
    

    Question

    prevnext

    A thief is stealing xots and yivs. Each xot has a mass of 18 kilograms and a volume of 6 liters. Each yiv has a mass of 9 kilograms and a volume of 12 liters. The thief can carry a maximum mass of 162 kilograms and a maximum volume of 72 liters. The profit from each xot is $6.85 and the profit from each yiv is $8.26.

    For your convenience, those numbers are organized in the table below.

    attribute	xot	yiv	capacity
    mass (kg)	18	9	162
    volume (L)	6	12	72
    profit ($)	6.85	8.26	\(\infty\)

    What is the maximum profit the thief can produce? (In dollars.)

    
    

    Solution

    Show solution

    This is an example of linear programming.

    Define \(x\) as number of xots and \(y\) as number of yivs. Write the constraints as inequalities.

    • Mass constraint: \(18x+9y \le 162\)
    • Volume constraint: \(6x+12y \le 72\)

    We can assume that optimal profit will occur at a boundary, so we can just consider the equations. \[18x+9y = 162\] \[6x+12y = 72\]

    If either equation has coefficients that all share a common factor, we can simplify by dividing both sides by the greatest common factor:

    \[2x+1y = 18\] \[1x+2y = 12\]

    We need to find the \(x\) intercepts and the \(y\) intercepts.

    To find the \(x\)-intercepts, set \(y\) to 0. \[2x_{\text{int}1}+(1)(0) = 18 \] \[2x_{\text{int}1} = 18 \] \[x_{\text{int}1} = \frac{18}{2} = 9 \]

    \[1x_{\text{int}2}+(2)(0) = 12 \] \[1x_{\text{int}2} = 12 \] \[x_{\text{int}2} = \frac{12}{1} = 12 \]

    To find the \(y\)-intercepts, set \(x\) to 0. \[(2)(0)+(1)(y_{\text{int1}}) = 18 \] \[1y_{\text{int}1} = 18 \] \[y_{\text{int}1} = \frac{18}{1} = 18 \]

    \[(1)(0)+(2)y_{\text{int2}} = 12 \] \[2y_{\text{int}2} = 12 \] \[y_{\text{int}2} = \frac{12}{2} = 6 \]

    To find the intersection, I would recommend elimination. Multiply the equations to have matching coefficients for \(y\) terms.

    \[4x + 2y = 36\] \[1x + 2y = 12\] Subtract equations.

    \[3x + (0)y = 24\]

    Simplify.

    \[3x = 24\]

    Divide by 3.

    \[x = 8\]

    Plug \(x = 8\) into the first equation. Solve for \(y\).

    \[(2)(8)+1y = 18\] \[16+1y = 18\] \[1y = 2\] \[y = 2\]

    So, the intersection occurs at \((8,\,2)\).

    You can plot the boundaries and shade the feasible region.

    
    

    The maximum profit could occur at \((0,6)\) or \((8,2)\) or \((9,0)\).

    Find those profits, using \(P(x,y)=6.85x+8.26y\).

    \[P(0,6)=(6.85)(0)+(8.26)(6) = 49.56\] \[P(8,2)=(6.85)(8)+(8.26)(2) = 71.32\] \[P(9,0)=(6.85)(9)+(8.26)(0) = 61.65\]

    So, the maximum profit is $71.32.

    It might help to visualize this in 3D, with xots, yivs, and profit on the 3 axes. Sorry, I was not able to make a 3D plot show up here.

    
    
76. 
    

    Question

    prevnext

    A thief is stealing xots and yivs. Each xot has a mass of 4 kilograms and a volume of 18 liters. Each yiv has a mass of 20 kilograms and a volume of 6 liters. The thief can carry a maximum mass of 80 kilograms and a maximum volume of 108 liters. The profit from each xot is $3.54 and the profit from each yiv is $2.41.

    For your convenience, those numbers are organized in the table below.

    attribute	xot	yiv	capacity
    mass (kg)	4	20	80
    volume (L)	18	6	108
    profit ($)	3.54	2.41	\(\infty\)

    What is the maximum profit the thief can produce? (In dollars.)

    
    

    Solution

    Show solution

    This is an example of linear programming.

    Define \(x\) as number of xots and \(y\) as number of yivs. Write the constraints as inequalities.

    • Mass constraint: \(4x+20y \le 80\)
    • Volume constraint: \(18x+6y \le 108\)

    We can assume that optimal profit will occur at a boundary, so we can just consider the equations. \[4x+20y = 80\] \[18x+6y = 108\]

    If either equation has coefficients that all share a common factor, we can simplify by dividing both sides by the greatest common factor:

    \[1x+5y = 20\] \[3x+1y = 18\]

    We need to find the \(x\) intercepts and the \(y\) intercepts.

    To find the \(x\)-intercepts, set \(y\) to 0. \[1x_{\text{int}1}+(5)(0) = 20 \] \[1x_{\text{int}1} = 20 \] \[x_{\text{int}1} = \frac{20}{1} = 20 \]

    \[3x_{\text{int}2}+(1)(0) = 18 \] \[3x_{\text{int}2} = 18 \] \[x_{\text{int}2} = \frac{18}{3} = 6 \]

    To find the \(y\)-intercepts, set \(x\) to 0. \[(1)(0)+(5)(y_{\text{int1}}) = 20 \] \[5y_{\text{int}1} = 20 \] \[y_{\text{int}1} = \frac{20}{5} = 4 \]

    \[(3)(0)+(1)y_{\text{int2}} = 18 \] \[1y_{\text{int}2} = 18 \] \[y_{\text{int}2} = \frac{18}{1} = 18 \]

    To find the intersection, I would recommend elimination. Multiply the equations to have matching coefficients for \(y\) terms.

    \[1x + 5y = 20\] \[15x + 5y = 90\] Subtract equations.

    \[14x + (0)y = 70\]

    Simplify.

    \[14x = 70\]

    Divide by 14.

    \[x = 5\]

    Plug \(x = 5\) into the first equation. Solve for \(y\).

    \[(1)(5)+5y = 20\] \[5+5y = 20\] \[5y = 15\] \[y = 3\]

    So, the intersection occurs at \((5,\,3)\).

    You can plot the boundaries and shade the feasible region.

    
    

    The maximum profit could occur at \((0,4)\) or \((5,3)\) or \((6,0)\).

    Find those profits, using \(P(x,y)=3.54x+2.41y\).

    \[P(0,4)=(3.54)(0)+(2.41)(4) = 9.64\] \[P(5,3)=(3.54)(5)+(2.41)(3) = 24.93\] \[P(6,0)=(3.54)(6)+(2.41)(0) = 21.24\]

    So, the maximum profit is $24.93.

    It might help to visualize this in 3D, with xots, yivs, and profit on the 3 axes. Sorry, I was not able to make a 3D plot show up here.

    
    
77. 
    

    Question

    prevnext

    A thief is stealing xots and yivs. Each xot has a mass of 6 kilograms and a volume of 4 liters. Each yiv has a mass of 9 kilograms and a volume of 12 liters. The thief can carry a maximum mass of 54 kilograms and a maximum volume of 48 liters. The profit from each xot is $1.55 and the profit from each yiv is $3.07.

    For your convenience, those numbers are organized in the table below.

    attribute	xot	yiv	capacity
    mass (kg)	6	9	54
    volume (L)	4	12	48
    profit ($)	1.55	3.07	\(\infty\)

    What is the maximum profit the thief can produce? (In dollars.)

    
    

    Solution

    Show solution

    This is an example of linear programming.

    Define \(x\) as number of xots and \(y\) as number of yivs. Write the constraints as inequalities.

    • Mass constraint: \(6x+9y \le 54\)
    • Volume constraint: \(4x+12y \le 48\)

    We can assume that optimal profit will occur at a boundary, so we can just consider the equations. \[6x+9y = 54\] \[4x+12y = 48\]

    If either equation has coefficients that all share a common factor, we can simplify by dividing both sides by the greatest common factor:

    \[2x+3y = 18\] \[1x+3y = 12\]

    We need to find the \(x\) intercepts and the \(y\) intercepts.

    To find the \(x\)-intercepts, set \(y\) to 0. \[2x_{\text{int}1}+(3)(0) = 18 \] \[2x_{\text{int}1} = 18 \] \[x_{\text{int}1} = \frac{18}{2} = 9 \]

    \[1x_{\text{int}2}+(3)(0) = 12 \] \[1x_{\text{int}2} = 12 \] \[x_{\text{int}2} = \frac{12}{1} = 12 \]

    To find the \(y\)-intercepts, set \(x\) to 0. \[(2)(0)+(3)(y_{\text{int1}}) = 18 \] \[3y_{\text{int}1} = 18 \] \[y_{\text{int}1} = \frac{18}{3} = 6 \]

    \[(1)(0)+(3)y_{\text{int2}} = 12 \] \[3y_{\text{int}2} = 12 \] \[y_{\text{int}2} = \frac{12}{3} = 4 \]

    To find the intersection, I would recommend elimination. Multiply the equations to have matching coefficients for \(y\) terms.

    \[2x + 3y = 18\] \[1x + 3y = 12\] Subtract equations.

    \[1x + (0)y = 6\]

    Simplify.

    \[1x = 6\]

    Divide by 1.

    \[x = 6\]

    Plug \(x = 6\) into the first equation. Solve for \(y\).

    \[(2)(6)+3y = 18\] \[12+3y = 18\] \[3y = 6\] \[y = 2\]

    So, the intersection occurs at \((6,\,2)\).

    You can plot the boundaries and shade the feasible region.

    
    

    The maximum profit could occur at \((0,4)\) or \((6,2)\) or \((9,0)\).

    Find those profits, using \(P(x,y)=1.55x+3.07y\).

    \[P(0,4)=(1.55)(0)+(3.07)(4) = 12.28\] \[P(6,2)=(1.55)(6)+(3.07)(2) = 15.44\] \[P(9,0)=(1.55)(9)+(3.07)(0) = 13.95\]

    So, the maximum profit is $15.44.

    It might help to visualize this in 3D, with xots, yivs, and profit on the 3 axes. Sorry, I was not able to make a 3D plot show up here.

    
    
78. 
    

    Question

    prevnext

    A thief is stealing xots and yivs. Each xot has a mass of 16 kilograms and a volume of 12 liters. Each yiv has a mass of 10 kilograms and a volume of 15 liters. The thief can carry a maximum mass of 160 kilograms and a maximum volume of 180 liters. The profit from each xot is $8.95 and the profit from each yiv is $7.86.

    For your convenience, those numbers are organized in the table below.

    attribute	xot	yiv	capacity
    mass (kg)	16	10	160
    volume (L)	12	15	180
    profit ($)	8.95	7.86	\(\infty\)

    What is the maximum profit the thief can produce? (In dollars.)

    
    

    Solution

    Show solution

    This is an example of linear programming.

    Define \(x\) as number of xots and \(y\) as number of yivs. Write the constraints as inequalities.

    • Mass constraint: \(16x+10y \le 160\)
    • Volume constraint: \(12x+15y \le 180\)

    We can assume that optimal profit will occur at a boundary, so we can just consider the equations. \[16x+10y = 160\] \[12x+15y = 180\]

    If either equation has coefficients that all share a common factor, we can simplify by dividing both sides by the greatest common factor:

    \[8x+5y = 80\] \[4x+5y = 60\]

    We need to find the \(x\) intercepts and the \(y\) intercepts.

    To find the \(x\)-intercepts, set \(y\) to 0. \[8x_{\text{int}1}+(5)(0) = 80 \] \[8x_{\text{int}1} = 80 \] \[x_{\text{int}1} = \frac{80}{8} = 10 \]

    \[4x_{\text{int}2}+(5)(0) = 60 \] \[4x_{\text{int}2} = 60 \] \[x_{\text{int}2} = \frac{60}{4} = 15 \]

    To find the \(y\)-intercepts, set \(x\) to 0. \[(8)(0)+(5)(y_{\text{int1}}) = 80 \] \[5y_{\text{int}1} = 80 \] \[y_{\text{int}1} = \frac{80}{5} = 16 \]

    \[(4)(0)+(5)y_{\text{int2}} = 60 \] \[5y_{\text{int}2} = 60 \] \[y_{\text{int}2} = \frac{60}{5} = 12 \]

    To find the intersection, I would recommend elimination. Multiply the equations to have matching coefficients for \(y\) terms.

    \[8x + 5y = 80\] \[4x + 5y = 60\] Subtract equations.

    \[4x + (0)y = 20\]

    Simplify.

    \[4x = 20\]

    Divide by 4.

    \[x = 5\]

    Plug \(x = 5\) into the first equation. Solve for \(y\).

    \[(8)(5)+5y = 80\] \[40+5y = 80\] \[5y = 40\] \[y = 8\]

    So, the intersection occurs at \((5,\,8)\).

    You can plot the boundaries and shade the feasible region.

    
    

    The maximum profit could occur at \((0,12)\) or \((5,8)\) or \((10,0)\).

    Find those profits, using \(P(x,y)=8.95x+7.86y\).

    \[P(0,12)=(8.95)(0)+(7.86)(12) = 94.32\] \[P(5,8)=(8.95)(5)+(7.86)(8) = 107.63\] \[P(10,0)=(8.95)(10)+(7.86)(0) = 89.5\]

    So, the maximum profit is $107.63.

    It might help to visualize this in 3D, with xots, yivs, and profit on the 3 axes. Sorry, I was not able to make a 3D plot show up here.

    
    
79. 
    

    Question

    prevnext

    A thief is stealing xots and yivs. Each xot has a mass of 16 kilograms and a volume of 5 liters. Each yiv has a mass of 4 kilograms and a volume of 15 liters. The thief can carry a maximum mass of 64 kilograms and a maximum volume of 75 liters. The profit from each xot is $5.68 and the profit from each yiv is $7.5.

    For your convenience, those numbers are organized in the table below.

    attribute	xot	yiv	capacity
    mass (kg)	16	4	64
    volume (L)	5	15	75
    profit ($)	5.68	7.50	\(\infty\)

    What is the maximum profit the thief can produce? (In dollars.)

    
    

    Solution

    Show solution

    This is an example of linear programming.

    Define \(x\) as number of xots and \(y\) as number of yivs. Write the constraints as inequalities.

    • Mass constraint: \(16x+4y \le 64\)
    • Volume constraint: \(5x+15y \le 75\)

    We can assume that optimal profit will occur at a boundary, so we can just consider the equations. \[16x+4y = 64\] \[5x+15y = 75\]

    If either equation has coefficients that all share a common factor, we can simplify by dividing both sides by the greatest common factor:

    \[4x+1y = 16\] \[1x+3y = 15\]

    We need to find the \(x\) intercepts and the \(y\) intercepts.

    To find the \(x\)-intercepts, set \(y\) to 0. \[4x_{\text{int}1}+(1)(0) = 16 \] \[4x_{\text{int}1} = 16 \] \[x_{\text{int}1} = \frac{16}{4} = 4 \]

    \[1x_{\text{int}2}+(3)(0) = 15 \] \[1x_{\text{int}2} = 15 \] \[x_{\text{int}2} = \frac{15}{1} = 15 \]

    To find the \(y\)-intercepts, set \(x\) to 0. \[(4)(0)+(1)(y_{\text{int1}}) = 16 \] \[1y_{\text{int}1} = 16 \] \[y_{\text{int}1} = \frac{16}{1} = 16 \]

    \[(1)(0)+(3)y_{\text{int2}} = 15 \] \[3y_{\text{int}2} = 15 \] \[y_{\text{int}2} = \frac{15}{3} = 5 \]

    To find the intersection, I would recommend elimination. Multiply the equations to have matching coefficients for \(y\) terms.

    \[12x + 3y = 48\] \[1x + 3y = 15\] Subtract equations.

    \[11x + (0)y = 33\]

    Simplify.

    \[11x = 33\]

    Divide by 11.

    \[x = 3\]

    Plug \(x = 3\) into the first equation. Solve for \(y\).

    \[(4)(3)+1y = 16\] \[12+1y = 16\] \[1y = 4\] \[y = 4\]

    So, the intersection occurs at \((3,\,4)\).

    You can plot the boundaries and shade the feasible region.

    
    

    The maximum profit could occur at \((0,5)\) or \((3,4)\) or \((4,0)\).

    Find those profits, using \(P(x,y)=5.68x+7.5y\).

    \[P(0,5)=(5.68)(0)+(7.5)(5) = 37.5\] \[P(3,4)=(5.68)(3)+(7.5)(4) = 47.04\] \[P(4,0)=(5.68)(4)+(7.5)(0) = 22.72\]

    So, the maximum profit is $47.04.

    It might help to visualize this in 3D, with xots, yivs, and profit on the 3 axes. Sorry, I was not able to make a 3D plot show up here.

    
    
80. 
    

    Question

    prevnext

    A thief is stealing xots and yivs. Each xot has a mass of 4 kilograms and a volume of 15 liters. Each yiv has a mass of 16 kilograms and a volume of 5 liters. The thief can carry a maximum mass of 64 kilograms and a maximum volume of 75 liters. The profit from each xot is $7.97 and the profit from each yiv is $3.62.

    For your convenience, those numbers are organized in the table below.

    attribute	xot	yiv	capacity
    mass (kg)	4	16	64
    volume (L)	15	5	75
    profit ($)	7.97	3.62	\(\infty\)

    What is the maximum profit the thief can produce? (In dollars.)

    
    

    Solution

    Show solution

    This is an example of linear programming.

    Define \(x\) as number of xots and \(y\) as number of yivs. Write the constraints as inequalities.

    • Mass constraint: \(4x+16y \le 64\)
    • Volume constraint: \(15x+5y \le 75\)

    We can assume that optimal profit will occur at a boundary, so we can just consider the equations. \[4x+16y = 64\] \[15x+5y = 75\]

    If either equation has coefficients that all share a common factor, we can simplify by dividing both sides by the greatest common factor:

    \[1x+4y = 16\] \[3x+1y = 15\]

    We need to find the \(x\) intercepts and the \(y\) intercepts.

    To find the \(x\)-intercepts, set \(y\) to 0. \[1x_{\text{int}1}+(4)(0) = 16 \] \[1x_{\text{int}1} = 16 \] \[x_{\text{int}1} = \frac{16}{1} = 16 \]

    \[3x_{\text{int}2}+(1)(0) = 15 \] \[3x_{\text{int}2} = 15 \] \[x_{\text{int}2} = \frac{15}{3} = 5 \]

    To find the \(y\)-intercepts, set \(x\) to 0. \[(1)(0)+(4)(y_{\text{int1}}) = 16 \] \[4y_{\text{int}1} = 16 \] \[y_{\text{int}1} = \frac{16}{4} = 4 \]

    \[(3)(0)+(1)y_{\text{int2}} = 15 \] \[1y_{\text{int}2} = 15 \] \[y_{\text{int}2} = \frac{15}{1} = 15 \]

    To find the intersection, I would recommend elimination. Multiply the equations to have matching coefficients for \(y\) terms.

    \[1x + 4y = 16\] \[12x + 4y = 60\] Subtract equations.

    \[11x + (0)y = 44\]

    Simplify.

    \[11x = 44\]

    Divide by 11.

    \[x = 4\]

    Plug \(x = 4\) into the first equation. Solve for \(y\).

    \[(1)(4)+4y = 16\] \[4+4y = 16\] \[4y = 12\] \[y = 3\]

    So, the intersection occurs at \((4,\,3)\).

    You can plot the boundaries and shade the feasible region.

    
    

    The maximum profit could occur at \((0,4)\) or \((4,3)\) or \((5,0)\).

    Find those profits, using \(P(x,y)=7.97x+3.62y\).

    \[P(0,4)=(7.97)(0)+(3.62)(4) = 14.48\] \[P(4,3)=(7.97)(4)+(3.62)(3) = 42.74\] \[P(5,0)=(7.97)(5)+(3.62)(0) = 39.85\]

    So, the maximum profit is $42.74.

    It might help to visualize this in 3D, with xots, yivs, and profit on the 3 axes. Sorry, I was not able to make a 3D plot show up here.